Short Lecture Notes: Interpolation Theory

and Function Spaces

Helmut Abels

July 27, 2011

1 Introduction

In the following let K = R or K = C.

Denition 1.1 Let X0, X1 be Banach spaces over K. Then the pair (X0, X1)is called admissible, compatible or an interpolation couple if there is a Hausdortopological vector space Z such that X0, X1 → Z with continuous embeddings.

Lemma 1.2 Let (X0, X1) be an admissible pair of Banach spaces. ThenX0 ∩X1 normed by

‖x‖X0∩X1 = max(‖x‖X0 , ‖x‖X1), x ∈ X0 ∩X1,

and X0 +X1 normed by

‖x‖X0+X1 = infx=x0+x1,x0∈X0,x1∈X1

‖x0‖X0 + ‖x1‖X1 , x ∈ X0 +X1,

are Banach spaces.

Notation: In the following we write for simplicity T ∈ L(Xj, Yj), j = 0, 1,for the condition that T : X0 + X1 → Y0 + Y1 is a linear operator withT |Xj ∈ L(Xj, Yj) for j = 0 and j = 1.

Denition 1.3 Let (X0, X1) and (Y0, Y1) be admissible pairs of Banach spacesand let X, Y be Banach spaces.

1. X is called intermediate space with respect to (X0, X1) if X0 ∩ X1 →X → X0 +X1 with continuous embeddings.

1

2. X and Y are called interpolation spaces with respect to (X0, X1) and(Y0, Y1) if X and Y are intermediate spaces with respect to (X0, X1) and(Y0, Y1), respectively, and if

T ∈ L(Xj, Yj), j = 0, 1, ⇒ T |X ∈ L(X, Y )

for all T : X0 +X1 → Y0 + Y1 linear.

3. X is called interpolation space with respect to (X0, X1) if the previousconditions hold with X = Y , (X0, X1) = (Y0, Y1).

4. Interpolation spaces X and Y with respect to (X0, X1) and (Y0, Y1) arecalled of exponent θ ∈ [0, 1], if there is some C > 0 such that

‖T‖L(X,Y ) ≤ C‖T‖1−θL(X0,Y0)‖T‖

θL(X1,Y1) (1.1)

for all T ∈ L(Xj, Yj), j = 0, 1. If (1.1) holds with C = 1, then X and Yare called exact of exponent θ. Analogously, X is an (exact) interpolationspace of exponent θ if the previous conditions hold for X = Y , (X0, X1) =(Y0, Y1).

Examples 1.4 1. Let 0 < α < 1, k ∈ N0 and let

C0(Rn) = f : Rn → K continuous and bounded‖f‖C0(Rn) = sup

x∈Rn|f(x)|

Ck(Rn) =f : Rn → K k-times cont. di. : ∂αx f ∈ C0(Rn)∀|α| ≤ k

‖f‖Ck(Rn) = sup

|α|≤ksupx∈Rn|∂αx f(x)|

Cα(Rn) =f : Rn → K : ‖f‖Cα(Rn) <∞

‖f‖Cα(Rn) = sup

x∈Rn|f(x)|+ sup

x,y∈Rn,x 6=y

|f(x)− f(y)||x− y|α

Then Cα(Rn) is an intermediate space of C0(Rn) and C1(Rn). Moreprecisely, there is some C > 0 such that

‖f‖Cα(Rn) ≤ C‖f‖1−αC0(Rn)‖f‖

αC1(Rn) for all f ∈ C1(Rn).

2

2. For 1 ≤ p ≤ ∞ let Lp(U, µ) denote the usual Lebesgue space on a measurespace (U, µ). Then for any 1 ≤ p0, p1, p ≤ ∞ such that 1

p= 1−θ

p0+ θ

p1for

some θ ∈ [0, 1] the space Lp(U, µ) is an intermediate space of Lp0(U, µ)and Lp1(U, µ). This follows from the generalized Hölder inequality

‖f‖Lp(U,µ) ≤ ‖f‖1−θLp0 (U,µ)‖f‖

θLp1 (U,µ) for all f ∈ Lp0(U, µ) ∩ Lp1(U, µ),

which itself follows easily from the standard Hölder inequality.

Remark 1.5 The last two inequalities are special cases of (1.1) if one choosesYj = K. The following theorem shows that Lp(U, µ) in the latter example isan exact interpolation space with respect to (Lp0(U, µ), Lp1(U, µ)) of expo-nent θ.

THEOREM 1.6 (Riesz-Thorin)Let 1 ≤ p0, p1, q0, q1 ≤ ∞, θ ∈ [0, 1], let (U, µ) and (V, ν) be σ-nite measurespaces and let K = C.1 Then

T ∈ L(Lpj(U, µ), Lqj(V, ν)), j = 0, 1 ⇒ T |Lp(U,µ) ∈ L(Lp(U, µ), Lq(V, ν)),

where1

p=

1− θp0

+θ

p1

,1

q=

1− θq0

+θ

q1

.

Moreover, for every T ∈ L(Lpj(U, µ), Lqj(V, ν)), j = 0, 1,

‖T‖L(Lp,Lq) ≤ ‖T‖1−θL(Lp0 ,Lq0 )‖T‖

θL(Lp1 ,Lq1 ).

The theorem will follow from the results of Section 2 below.

Remark 1.7 The proof of the Riesz-Thorin interpolation theorem providedthe fundamental ideas for the so-called complex interpolation method. Gener-ally an interpolation method is an functor F from the cathegory of admissiblecouples of Banach spaces to the cathegory of Banach spaces such that anypair of admissible couples X = (X0, X1), Y = (Y0, Y1) is mapped to aninterpolation space F(X, Y ) with respect to X, Y .2

1Meaning that all functions will be complex valued in the following.2Readers not familiar with the basic notion of cathegory do not have to understand

this precisely since no cathegory theory will be needed in the following.

3

In some situations the following theorem due to Marcinkiewicz can beapplied, when an application of the Riesz-Thorin theorem fails.

THEOREM 1.8 (Marcinkiewicz)Let 1 ≤ p0, p1, q0, q1 ≤ ∞ with q0 6= q1, let θ ∈ (0, 1), and let (U, µ) and (V, ν)be measure spaces. Moreover, let

1

p=

1− θp0

+θ

p1

,1

q=

1− θq0

+θ

q1

and assume that p ≤ q. Then

T ∈ L(Lpj(U, µ), Lqj∗ (V, ν)), j = 0, 1 ⇒ T |Lp(U,µ) ∈ L(Lp(U, µ), Lq(V, ν)),

Moreover, there is some Cθ such that for every T ∈ L(Lpj(U, µ), Lqj∗ (V, ν)), j =

0, 1,‖T‖L(Lp,Lq) ≤ Cθ‖T‖1−θ

L(Lp0 ,Lq0∗ )‖T‖θL(Lp1 ,L

q1∗ ).

Here Lq∗(V, ν) is the weak Lq-space, dened by

Lq∗(V, ν) =

f : V → K measurable : m(t; f) ≤ C

tqfor all t > 0 and some C > 0

if 1 ≤ q <∞, where

m(t; f) = ν(x ∈ V : |f(x)| > t), t > 0,

denotes the distribution function of f . Moreover, if q =∞, then L∞∗ (V, ν) =L∞(V, ν). We note that Lq∗(V, ν) is quasi-normed space with quasi-norm by

‖f‖Lq∗ = supt>0

tm(t; f)1q .

Here a quasi-norm ‖.‖ on a vector space X satises the same conditions asa norm except that the triangle inequality is replaced by

‖x+ y‖ ≤ C(‖x‖+ ‖y‖) for all x, y ∈ X

for some C ≥ 1. The bounded operators and the operator norm is dened inthe same way for quasi-normed spaces as for normed spaces.Remark: One also denotes Lq,∞(V, ν) ≡ Lq∗(V, ν).

4

A proof of the Marcinkiewicz interpolation theorem in the special casep0 = q0 = 1, p1 = q1 = r, (U, µ) = (Rn, λn) can be found in the appendix. Theproof provided the basic ideas for the so-called real interpolation method. Thereal and the complex interpolation methods are the most important methodsto construct interpolation spaces and will be one of the central topic of thislecture series. We will start with the complex method since it is easier tounderstand in the beginning, although the real interpolation method is insome sense more fexible and universal.

2 Complex Interpolation Method

2.1 Holomorphic Functions on the Strip

For the following let X be a complex Banach space and let S = z ∈ C : 0 ≤Re z ≤ 1, S0 = z ∈ C : 0 < Re z < 1. A mapping f : S0 → X is calledholomorphic if z 7→ 〈f(z), x′〉, z ∈ S0, is holomorphic in the usual sense forall x′ ∈ X ′.

THEOREM 2.1 (Maximum Principle/Phragmen-Lindelöf Theorem)Let f : S → X be continuous, bounded, and holomorphic in the interior of S.Then

supz∈S‖f(z)‖X ≤ max

(supt∈R‖f(it)‖X , sup

t∈R‖f(1 + it)‖X

). (2.1)

Proof: First let X = C and assume that f(z) →| Im z|→∞ 0. Consider themapping h : S → C with

h(z) =eiπz − ieiπz + i

, z ∈ S. (2.2)

Then h is a bijective mapping from S onto U = z ∈ C : |z| ≤ 1 \ ±1,that is holomorphic in S0 and maps ∂S onto |z| = 1 \ ±1. Thereforeg(z) := f(h−1(z)) is bounded and continuous on U and analytic in the interiorof U . Moreover, because of lim| Im z|→∞ f(z) = 0, limz→±1 g(z) = 0 and wecan extend g to a continuous function on |z| ≤ 1. Hence

|g(z)| ≤ max|w|=1|g(w)| = max

(supt∈R|f(it)|, sup

t∈R|f(1 + it)|

),

5

which implies the statement in this case. Next, if X = C and f is a generalfunction as in the assumptions, then we consider fδ,z0(z) = eδ(z−z0)2f(z),δ > 0, z0 ∈ S0. Since |eδ(z−z0)2| ≤ eδ(x

2−y2) with z − z0 = x + iy, −1 ≤ x ≤1, y ∈ R, we have fδ,z0(z)→| Im z|→∞ 0. Therefore

|f(z0)| = |fδ,z0(z0)| ≤ max

(supt∈R|fδ,z0(it)|, sup

t∈R|fδ,z0(1 + it)|

)≤ eδ max

(supt∈R|f(it)|, sup

t∈R|f(1 + it)|

).

Passing to the limit δ → 0, we obtain (2.1) in the case X = C since z0 ∈ Swas arbitrary.

Finally, if f : S → X is as in the assumption and z0 ∈ S0, we can ndsome x′ ∈ X ′ with ‖x′‖X′ = 1 such that ‖f(z0)‖X = |〈f(z0), x′〉|. Now oneapplies (2.1) to gx′(z) := 〈f(z), x′〉 and obtains

‖f(z0)‖X = |gx′(z0)| ≤ max

(supt∈R|gx′(it)|, sup

t∈R|gx′(1 + it)|

)≤ max

(supt∈R‖f(it)‖X , sup

t∈R‖f(1 + it)‖X

).

for arbitrary z0 ∈ S0.As a corollary we obtain the following three line theorem, which is the basisfor the proof of the Riesz-Thorin interpolation theorem and the complexinterpolation method.

THEOREM 2.2 (Three Lines Theorem) Let f : S → X be continuous,bounded, and holomorphic in the interior of S. Then

supt∈R‖f(θ + it)‖X ≤

(supt∈R‖f(it)‖X

)1−θ (supt∈R‖f(1 + it)‖X

)θ. (2.3)

for every θ ∈ [0, 1].

Proof: Let M0 = supt∈R ‖f(it)‖X , M1 = supt∈R ‖f(1 + it)‖X . Let λ ∈ Rand dene

Fλ(z) = eλzf(z).

Then‖Fλ(z)‖X ≤ max(M0, e

λM1)

6

by Theorem 2.1. Hence

‖f(θ + it)‖X ≤ max(e−λθM0, e

λ(1−θ)M1

)for all t ∈ R. Choosing λ = ln M0

M1nishes the proof.

2.2 Complex Interpolation Method and Lp-Spaces

Let (X0, X1) be an admissible pair of complex Banach spaces. In particularwe will assume always K = C in the following!

Denition 2.3 Let F (X0, X1) be the set of all f : S → X0 + X1 that arecontinuous, bounded, and holomorphic in S0 and such that

t 7→ f(it) ∈ C(R;X0), t 7→ f(1 + it) ∈ C(R;X1).

We equip F (X0, X1) with the norm

‖f‖F (X0,X1) = max

(supt∈R‖f(it)‖X0 , sup

t∈R‖f(1 + it)‖X1

).

Finally, let F0(X0, X1) be the set of all f ∈ F (X0, X1) such that

lim|t|→∞

‖f(j + it)‖Xj = 0 for j = 0, 1.

Remark 2.4 Because of Theorem 2.1, we have F (X0, X1) → C(S;X0+X1).Using this, it is not dicult to show that F (X0, X1) is a Banach space.Moreover, obviously F0(X0, X1) is a closed subspace of F (X0, X1).

We will need the following technical lemma.

Lemma 2.5 The linear hull of the functions eδz2+λza, a ∈ X0 ∩X1, δ > 0,

λ ∈ R is dense in F0(X0, X1).

We refer to [1, Lemma 4.2.3] for a proof.

Denition 2.6 Let θ ∈ (0, 1). Then the complex interpolation space (X0, X1)[θ]

is dened as

(X0, X1)[θ] = f(θ) : f ∈ F (X0, X1)‖x‖(X0,X1)[θ] = inf

f∈F (X0,X1):f(θ)=x‖f‖F (X0,X1).

7

Remarks 2.7 1. Since Nθ = f ∈ F (X0, X1) : f(θ) = 0 is closed inF (X0, X1), we have that (X0, X1)[θ]

∼= F (X0, X1)/Nθ is a Banach space.

2. For every θ ∈ (0, 1) we have

(X0, X1)[θ] = (X1, X0)[1−θ].

3. One can replace F (X0, X1) by F0(X0, X1) in the denition of (X0, X1)[θ].In order to see that this gives the same space and norm, one uses thatfδ = eδ(z−θ)

2f(z) ∈ F0(X0, X1) if f ∈ F (X0, X1) and δ > 0. Moreover,

fδ(θ) = f(θ) and ‖fδ‖F (X0,X1) ≤ max(eδθ2, eδ(1−θ)

2)‖f‖F (X0,X1).

4. We have X0 ∩ X1 → (X0, X1)[θ] since we can choose f(z) = x for allz ∈ S and x ∈ X0 ∩ X1. Moreover, (X0, X1)[θ] → X0 + X1 sinceF (X0, X1) → C(S;X0 + X1). Summing up, we have (X0, X1)[θ] is anintermediate space with respect to (X0, X1), i.e.,

X0 ∩X1 → (X0, X1)[θ] → X0 +X1

5. For every θ ∈ (0, 1) we have (X0, X0)[θ] = X0 with same norm.

Because of Lemma 2.5 and statement 3 above, we obtain

Corollary 2.8 Let θ ∈ (0, 1). Then X0 ∩X1 is dense in (X0, X1)[θ].

THEOREM 2.9 Let 1 ≤ p0, p1 ≤ ∞, θ ∈ (0, 1), and let (U, µ) be a measurespace. Then

(Lp0(U, µ), Lp1(U, µ))[θ] = Lp(U, µ)

with the same norm, where

1

p=

1− θp0

+θ

p1

.

Proof: See [2, Example 2.1.11].

8

2.3 Properties of Complex Interpolation Spaces

THEOREM 2.10 Let θ ∈ (0, 1) and let (X0, X1) and (Y0, Y1) be two ad-missible pairs of Banach spaces. Then (X0, X1)[θ] and (Y0, Y1)[θ] are exactinterpolation spaces of exponent θ with respect to (X0, X1) and (Y0, Y1). Inparticular,

‖T‖L(Xθ,Yθ) ≤ ‖T‖1−θL(X0,Y0)‖T‖

θL(X1,Y1)

for every T ∈ L(Xj, Yj), j = 0, 1, where Xθ = (X0, X1)[θ], Yθ = (Y0, Y1)[θ]

Proof: See [2, Theorem 2.1.6]

Corollary 2.11 For every θ ∈ (0, 1) we have

‖x‖(X0,X1)[θ] ≤ ‖x‖1−θX0‖x‖θX1

for all x ∈ X0 ∩X1.

Moreover, we note that Theorem 2.10 and Theorem 2.9 imply the Riesz-Thorin Interpolation Theorem 1.6.

2.4 Extensions and Generalizations

The proof of the Theorem 2.9 can be easily modied to the case of weighted-Lp-spaces with respect to a xed reference measure µ. More precisely:

THEOREM 2.12 Let 1 ≤ p0, p1 < ∞, θ ∈ (0, 1), and let (U, µ) be ameasure space. Moreover, let ωj : U → (0,∞), j = 0, 1, be measurable andset

ω(x) = ω0(x)pp0

(1−θ)ω1(x)

pp1θ.

Then(Lp0(U, ω0 dµ), Lp1(U, ω1 dµ))[θ] = Lp(U, ω dµ).

with the same norm, where

1

p=

1− θp0

+θ

p1

.

Here the measure ω dµ is dened by

ω dµ(A) =

∫A

ω(x) dµ(x)

for every measurable A ⊆ U .

9

A useful generalization of Theorem 2.10 is:

THEOREM 2.13 Let (X0, X1), (Y0, Y1) be admissible couples of Banachspaces and let Tz ∈ L(X0 ∩ X1, Y0 + Y1), z ∈ S, be such that z 7→ Tzx iscontinuous on S and holomorphic in S0 for every x ∈ X0 ∩ X1. Moreover,assume that t 7→ Tit ∈ C(R;L(X0, Y0)) and t 7→ T1+it ∈ C(R;L(X1, Y1))(with bounded operator norms). Then for every θ ∈ (0, 1)

‖Tθx‖(Y0,Y1)[θ] ≤M1−θ0 M θ

1‖x‖(X0,X1)[θ] for all x ∈ X0 ∩X1,

whereM0 = sup

t∈R‖Tit‖L(X0,Y0), M1 = sup

t∈R‖T1+it‖L(X1,Y1).

In particular Tθ extends to a bounded operator from (X0, X1)[θ] to (Y0, Y1)[θ]

with operator norm bounded by M1−θ0 M θ

1 .

Proof: See [2, Theorem 2.1.7].

10

3 Real Interpolation Method

3.1 K-Method

In the following let (X0, X1) be an admissible pair of Banach spaces.

Denition 3.1 For t > 0, x ∈ X0 +X1 let

K(t, x) ≡ K(t, x;X0, X1) = infx=x0+x1,x0∈X0,x1∈X1

‖x0‖X0 + t‖x1‖X1 .

Lemma 3.2 For all x ∈ X0 +X1 the mapping t 7→ K(t, x) is non-negative,increasing and concave. Moreover,

K(t, x) ≤ max

(1,t

s

)K(s, x) for all t, s > 0. (3.1)

Denition 3.3 For θ ∈ (0, 1), 1 ≤ p ≤ ∞ we dene the real interpolation

space (X0, X1)θ,p as

(X0, X1)θ,p := x ∈ X0 +X1 : Φθ,p(K(·, x)) <∞

where

Φθ,p(K(·, x)) =∥∥t−θK(t, x)

∥∥Lp((0,∞), dt

t)

=

(∫∞0

(t−θK(t, x)

)p dtt

) 1p if 1 ≤ p <∞,

supt>0 t−θK(t, x) if p =∞.

(X0, X1)θ,p is normed by ‖x‖θ,p := Φθ,p(K(·, x)). Moreover, for θ ∈ (0, 1)

(X0, X1)θ :=x ∈ (X0, X1)θ,∞ : lim

t→0t−θK(t, x) = lim

t→∞t−θK(t, x) = 0

denotes the continuous interpolation space of (X0, X1).

Remarks 3.4 1. It is easy to check that ‖·‖θ,p is indeed a norm and that(X0, X1)θ,p is a linear subspace of X0 + X1. Moreover, (X0, X1)θ is aclosed subspace of (X0, X1)θ,∞.

11

2. Important properties of the integral with respect to dtt are:∫ ∞

0

f(at)dt

t=

∫ ∞0

f(t)dt

t,

∫ ∞0

f(tα)dt

t=

1

|α|

∫ ∞0

f(t)dt

t

for any a, α 6= 0 and integrable function f .3

The main result of this section is:

THEOREM 3.5 Let 1 ≤ p ≤ ∞, θ ∈ (0, 1) and let (X0, X1), (Y0, Y1)be admissible pairs of Banach spaces. Then (X0, X1)θ,p and (Y0, Y1)θ,p areinterpolation spaces with respect to (X0, X1) and (Y0, Y1) that are exact ofexponent θ. Moreover, (X0, X1)θ and (Y0, Y1)θ are interpolation spaces withrespect to (X0, X1) and (Y0, Y1) that are exact of exponent θ

The theorem follows from the following lemmas:

Lemma 3.6 Let 1 ≤ p1 ≤ p2 ≤ ∞, θ ∈ (0, 1). Then

X0 ∩X1 → (X0, X1)θ,p1 → (X0, X1)θ,p2 → X0 +X1 (3.2)

with continuous embeddings. Moreover, (X0, X1)θ,p → (X0, X1)θ for any1 ≤ p <∞. In particular, (X0, X1)θ,p and (X0, X1)θ are intermediate spaceswith respect to (X0, X1).

Proof: See [2, Proposition 1.1.3].

Lemma 3.7 Let the assumptions of Theorem 3.5 be valid and let X =(X0, X1)θ,p, Y = (Y0, Y1)θ,p. Then

T ∈ L(Xj, Yj), j = 0, 1, ⇒ T |X ∈ L(X, Y )

and‖T‖L(X,Y ) ≤ ‖T‖1−θ

L(X0,Y0)‖T‖θL(X1,Y1)

for all T ∈ L(Xj, Yj), j = 0, 1. The same is true for X = (X0, X1)θ andY = (Y0, Y1)θ.

3The rst property reects the fact that f is a Haar measure on the topological group

((0,∞), ·). I.e. it is a regular Borel measure that is invariant with respect to group action

f 7→ f(a · .).

12

Proof: See [2, Theorem 1.1.6].

Note that the constants/the operator norms of the embeddings in (3.2)depend on (θ, p).

Lemma 3.8 For any θ ∈ (0, 1), 1 ≤ p ≤ ∞ the normed spaces (X0, X1)θ,pand (X0, X1)θ are complete.

Proof: See [2, Proposition 1.1.3].

Some more properties are as follows:

1. Since K(t, x;X0, X1) = tK(t−1, x;X1, X0), we have

(X0, X1)θ,p = (X1, X0)1−θ,p (X0, X1)θ = (X1, X0)1−θ

for all θ ∈ (0, 1), 1 ≤ p ≤ ∞.

2. If X1 → X0 continuously, then K(t, x) ≤ ‖x‖X0 and therefore

cθ,p‖x‖θ,p ≤ ‖x‖X0 + ‖t−θK(t, x;X0;X1)‖Lp((0,1); dtt

) ≤ Cθ,p‖x‖θ,p

for some constants cθ,p, Cθ,p > 0 depending on θ ∈ (0, 1), 1 ≤ p ≤ ∞.

3. For all θ ∈ (0, 1), 1 ≤ p ≤ ∞ we have (X0, X0)θ,p = X0 with equivalentnorms. More precisely,

cθ,p‖x‖θ,p ≤ ‖x‖X0 ≤ Cθ,p‖x‖θ,p

for some constants Cθ,p, cθ,p > 0 depending on θ and p!!!

Proposition 3.9 Let X1 → X0 continuously. Then for every 0 < θ1 < θ2 < 1we have

(X0, X1)θ2,∞ → (X0, X1)θ1,1.

Recall that, if 1 ≤ p ≤ ∞ and Ω ⊆ Rn is a domain, then

W 1p (Ω) = f ∈ Lp(Ω) : ∇f ∈ Lp(Ω) ,

‖f‖W 1p (Ω) = ‖f‖Lp(Ω) + ‖∇f‖Lp(Ω).

13

Here ∇f = (∂x1f, . . . , ∂xnf) denotes the weak or distributional derivatives off . More precisely, gj ∈ Lp(Ω) for 1 ≤ j ≤ n is called weak derivative of f indirection xj, if ∫

Ω

gj(x)ϕ(x) dx = −∫

Ω

f(x)∂xjϕ(x) dx

for all smooth and compactly supported ϕ : Ω→ K, i.e., for all ϕ ∈ C∞0 (Ω). Ifthe weak derivative exists, it is unique and it will be for simplicity be denotedby ∂xjf . Note that, if ψ ∈ C∞0 (Rn) and f ∈ W 1

p (Rn), then ψ ∗ f ∈ C1(Rn)and

∂xjψ ∗ f(x) = −∫Rn∂yjψ(x− y)f(y) dy =

∫Rnψ(x− y)∂yjf(y) dy

because of the denition of the weak derivative. Choosing e.g. ψ(x) =ϕε(x) = ε−nϕ(x

ε), ε > 0, where ϕ ∈ C∞0 (Rn) with

∫Rn ϕ(x) dx = 1, one

obtains that ϕε ∗ f →ε→0 f in W 1p (Rn). Hence C∞(Rn)∩W 1

p (Rn) is dense inW 1p (Rn).

Moreover, if U ⊆ Rn is an open set, then Cθ(U), θ ∈ (0, 1), denotes theset of all bounded and Hölder continuous functions f : U → K of degree θsuch that

‖f‖Cθ(U) = ‖f‖∞ + supx 6=y,x,y∈U

|f(x)− f(y)||x− y|θ

Example 3.10 For every 0 < θ < 1 and 1 ≤ p <∞ we have

(C(Rn), C1(Rn))θ,∞ = Cθ(Rn), (3.3)

(C(Rn), C1(Rn))θ = hθ(Rn), (3.4)

(Lp(Rn),W 1p (Rn))θ,p = W θ

p (Rn), (3.5)

with equivalent norms, where hθ(Rn) is the little Hölder space dened by

hθ(Rn) =

f ∈ Cθ(Rn) : lim

R→0+sup

x,y∈Rn,0<|x−y|≤R

|f(x)− f(y)||x− y|θ

= 0

equipped with the norm of Cθ(Rn) and W θ

p (Rn) is the Sobolev-Slobodetskii

space dened by

W θp (Rn) =

f ∈ Lp(Rn) : [f ]W θ

p (Rn) <∞,

[f ]W θp (Rn) =

(∫Rn×Rn

|f(x)− f(y)|p

|x− y|θp+nd(x, y)

) 1p

.

14

Example 3.11 Let Ω ⊂ Rn be a domain such that there is an extension oper-

ator E ∈ L(Cθ(Ω);Cθ(Rn)) for all θ ∈ [0, 1] and Ef |Ω = f on Ω. Then

(C(Ω), C1(Ω))θ,∞ = Cθ(Ω)

for all 0 < θ < 1.

Remark 3.12 If Ω = Rn+ = Rn−1 × (0,∞), then E dened by

Ef(x) =

f(x) if xn ≥ 0

3f(x′,−xn)− 2f(x′,−2xn) if xn < 0

where x′ = (x1, . . . , xn−1) satises the conditions of (3.11). More generally,every bounded domain Ω with C1-boundary satises the conditions of (3.11).This can be shown by locally mapping the boundary of Ω to pieces of a halfspace Rn

+ and applying the half-space extension operator together with asuitable partition of unity.

Remark 3.13 On the basis on (3.5), we will be able to prove that

Trxn=0 W1p (Rn

+) = (Lp(Rn−1),W 1p (Rn−1))1− 1

p,p,

where 1 < p <∞, Rn+ = Rn−1 × (0,∞), and

Trxn=0W1p (Rn

+) =a ∈ Lp(Rn−1) : a(x′) = f(x′, 0) for some f ∈ W 1

p (Rn+)

equipped with the quotient norm

‖a‖Trxn=0W 1p (Rn+) = inf

f∈W 1p (Rn+):f |xn=0=a

‖f‖W 1p (Rn+)

To this end we will need an equivalent characterization of the real interpo-lation spaces (X0, X1)θ,p, namely the so-called trace method. But before weneed some preliminaries on vector-valued Sobolev spaces.

3.2 Excursion: Vector-Valued Lp and Sobolev Spaces

Denition 3.14 Let X be a Banach space and let (M,µ) be a measure space.Then a mapping f : M → X is called simple if f =

∑Nk=1 αkχAk for some

measurable Ak ⊆ M and αk ∈ X. Moreover, f : M → X is called strongly

15

measurable if there is a sequence fk : M → X of simple functions fk, k ∈ N,such that fk(x) →k→∞ f(x) in X for almost every x ∈ M . Moreover, f iscalled Bochner integrable if there is a sequence fk : M → X, k ∈ N, of simplefunctions such that

limk→∞

∫M

‖fk(x)− f(x)‖X dµ(x) = 0. (3.6)

If f =∑N

k=1 αkχAk is a simple function, then the X-valued integral is denedby ∫

M

f(x) dµ(x) :=N∑k=1

αkµ(Ak).

If f is Bochner integrable and fk are simple functions such that (3.6) holds,then

(∫Mfk(x) dµ(x)

)k∈N is a Cauchy sequence in X and we dene∫

M

f(x) dµ(x) = limk→∞

∫M

fk(x) dµ(x).

As in the scalar case the denition is independent of the sequence fk.A useful criterion is the following:

Proposition 3.15 Let f : M → X. Then f is Bochner integrable if and onlyif f is strongly measurable and x 7→ ‖f(x)‖X is integrable.

We refer to the book by Yosida, Functional Analysis, Theorem 1 in ChapterV, Section 5.

From the denitions it easily follows that, if f : M → X is Bochnerintegrable and A ∈ L(X, Y ), then Af : M → Y is Bochner integrable (w.r.t.Y ) and

A

∫M

f(x) dµ(x) =

∫M

Af(x) dµ(x).

In particular, if w′ ∈ X ′ and f : M → X is Bochner integrable, then x 7→〈f(x), w′〉 is integrable and⟨∫

M

f(x) dµ(x), w′⟩

=

∫M

〈f(x), w′〉 dµ(x).

Here 〈w,w′〉 = w′(w) denotes the duality product of w ∈ X and w ∈ X ′.Similarly to the scalar case we dene for 1 ≤ p ≤ ∞

Lp(M ;X) = f : M → X strongly measurable : ‖f‖X ∈ Lp(M)‖f‖Lp(M ;X) = ‖‖f(.)‖X‖Lp(M)

16

In the following we restrict ourselves for simplicity to an interval (a, b) to-gether with the Lebesgue measure.

Denition 3.16 Let f ∈ L1((a, b);X) ≡ L1(a, b;X). Then g ∈ L1(a, b;X)is called weak derivative if∫ b

a

g(t)ϕ(t) dt = −∫ b

a

f(t)ϕ′(t) dt

for all ϕ ∈ C∞0 ((a, b)).

Note that, if g ∈ L1(a, b;X) is a weak derivative of f ∈ L1(a, b;X), thengw(x) := 〈g(x), w〉 is a weak derivative of the scalar function fw(x) = 〈f(x), w〉for every w ∈ X ′. In particular, the weak derivative of f is unique if it exists.Therefore it will for simplicity be denoted by f ′. Finally, we set

W 1p (a, b;X) = f ∈ Lp(a, b;X) : f has a weak derivative f ′ ∈ Lp(a, b;X)

‖f‖W 1p (a,b;X) = ‖f‖Lp(a,b;X) + ‖f ′‖Lp(a,b;X)

Lemma 3.17 Let −∞ < a < b < ∞, X be a Banach space and let 1 ≤p < ∞. Then C1([a, b];X) is dense in W 1

p (a, b;X) and W 1p (a, b;X) →

C0([a, b];X) continuously. Moreover, for every f ∈ W 1p (a, b;X)

f(y)− f(x) =

∫ y

x

f ′(t) dt for all a ≤ x ≤ y ≤ b.

Here Ck([a, b];X), k ∈ N0, is the Banach space of all k-times dierentiablefunctions f : [a, b]→ X equipped with a standard norm, e.g.

‖f‖Ck([a,b];X) = supx∈[a,b],j=0,...,k

∥∥f (j)(x)∥∥X

Remark 3.18 If 1 < p <∞, we even have

W 1p (a, b;X) → C1− 1

p ([a, b];X).

where Cα([a, b];X), α ∈ [0, 1], is dened as in the scalar case just replacing|.| by ‖.‖X in the norms, i.e.,

‖f‖Cα([a,b];X) = ‖f‖C0([a,b];X) + supx 6=y,x,y∈[a,b]

‖f(x)− f(y)‖X|x− y|α

17

3.3 The Trace Method

Denition 3.19 Let (X0, X1) be an admissible pair of Banach spaces and letθ ∈ (0, 1), 1 ≤ p ≤ ∞. Then we dene V (p, θ,X1, X0) as the set of allu : (0,∞) → X0 + X1 such that u ∈ W 1

p (a, b;X0 + X1) ∩ Lp(a, b;X1) withu′ ∈ Lp(a, b;X0) for all 0 < a < b <∞ and nite ‖u‖V (p,θ,X1,X0) where

‖u‖V (p,θ,X0,X1) =

(∫∞0tθp (‖u(t)‖X1 + ‖u′(t)‖X0)

p dtt

) 1p if p <∞,

ess supt>0 tθ (‖u(t)‖X1 + ‖u′(t)‖X0) if p =∞.

Remark 3.20 If u ∈ V (p, θ,X1, X0), then limt→0 u(t) exists in X0 + X1

because of

‖u(t)− u(s)‖X0 ≤∫ t

s

‖u′(τ)‖X0 dτ =

∫ t

s

τ θ‖u′(τ)‖X0τ1−θ dτ

τ

≤(∫ t

s

‖τ θu′(τ)‖pX0

dτ

τ

) 1p(∫ t

s

τ (1−θ)p′ dτ

τ

) 1p′

≤ ‖u‖V (p,θ,X0,X1) (F (t)− F (s))1p′

for all 0 < s < t < ∞ in the case 1 < p < ∞, where t 7→ F (t) := t(1−θ),t ∈ [0,∞), is a continuous function. If p = 1 or p = ∞, one simply replacesthe Lp(0,∞; dt

t)-norm or Lp

′(0,∞; dt

t)-norm above by the L∞(0,∞; dt

t)-norm.

THEOREM 3.21 Let (X0, X1) be an admissible pair of Banach spaces,θ ∈ (0, 1) and 1 ≤ p ≤ ∞. Then

(X0, X1)θ,p = x ∈ X0 +X1 : x = u(0), u ∈ V (p, 1− θ,X1, X0) .

Moreover, the norm dened by

‖x‖Trθ,p = inf

‖u‖V (p,1−θ,X1,X0) : x = u(0), u ∈ V (p, 1− θ,X1, X0)

is equivalent to the norm of (X0, X1)θ,p.

For the proof we need the following Hardy-type inequalities:

Lemma 3.22 Let α > 0 and let 1 ≤ p < ∞. Then there is a constant Cα,psuch that ∫ ∞

0

t−αp(∫ t

0

ϕ(s)ds

s

)pdt

t≤ Cα,p

∫ ∞0

t−αpϕ(t)pdt

t(3.7)∫ ∞

0

tαp(∫ ∞

t

ϕ(s)ds

s

)pdt

t≤ Cα,p

∫ ∞0

tαpϕ(t)pdt

t(3.8)

18

for all non-negative measurable functions ϕ : (0,∞)→ R.

Remarks: It can be shown that Cα,p = |α|−p is the optimal constant in thelatter inequalities, cf. the book by Hardy et al. Inequalities, Theorem 330.

Choosing α = p−1p, ϕ(t) = t|f ′(t)| and using |f(t)| ≤

∫ t0|f ′(τ)| dτ in

Lemma 3.22, one easily derives the classical Hardy inequality:∥∥∥∥f(t)

t

∥∥∥∥Lp(0,∞)

≤ Cp‖f ′‖Lp(0,∞)

for all f ∈ C∞0 (0,∞), where 1 < p <∞.

Remark 3.23 The proof of Theorem 3.21 even shows that for x ∈ (X0, X1)θ,pthere some let v ∈ V (p, 1− θ,X1, X0) be such that v(0) = x and

t 7→ t2−θ‖v′(t)‖X1 ∈ Lp(

0,∞;dt

t

).

Moreover, there is some Cθ,p > 0 such that

‖v‖V (p,1−θ,X1,X0) +∥∥t2−θ‖v′(t)‖X1

∥∥Lp(0,∞; dt

t)≤ Cθ,p‖x‖θ,p.

Corollary 3.24 Let θ ∈ (0, 1) and let 1 ≤ p < ∞. Then X0 ∩X1 is densein (X0, X1)θ,p.

Corollary 3.25 Let 1 ≤ p < ∞ and let Rn+ = x ∈ Rn : xn > 0. Then

for every v ∈ W 1p (Rn

+) we have Trxn=0 v := v|xn=0 ∈ W1− 1

pp (Rn−1). Moreover,

the mapping Trxn=0 : W 1p (Rn

+)→ W1− 1

pp (Rn−1) is bounded, linear and onto.

Remark 3.26 Let

V0(∞, θ,X1, X0) =

u ∈ V (∞, θ,X1, X0) : lim

t→0/∞tθ (‖u(t)‖X1 + ‖u′(t)‖X0) = 0

.

Then

(X0, X1)θ = x ∈ X0 +X1 : x = u(0) for some u ∈ V0(∞, 1− θ,X1, X0) .

We refer to [2, Proposition 1.13] for a proof.4

4The proof is correct if X1 → X0 and can be modied for the general case.

19

3.4 Intermediate Spaces and Reiteration

Denition 3.27 Let θ ∈ [0, 1], let (X0, X1) be an admissible pair of Banachspaces and let Y be an intermediate space with respect to (X0, X1). Then

1. Y is said to belong to the class Jθ if there is a constant C > 0 such that

‖x‖Y ≤ C‖x‖1−θX0‖x‖θX1

for all x ∈ X0 ∩X1.

In this case we write Y ∈ Jθ(X0, X1).

2. Y is said to belong to the class Kθ if there is a constant k > 0 such that

K(t, x) ≤ ktθ‖x‖Y for all x ∈ Y.

In this case we write Y ∈ Kθ(X0, X1).

Note that, if θ ∈ (0, 1), then Y ∈ Kθ(X0, X1) means that Y → (X0, X1)θ,∞continuously.

Proposition 3.28 Let θ ∈ (0, 1) and let Y be an intermediate space with re-spect to an admissible couple of Banach spaces (X0, X1). Then Y ∈ Jθ(X0, X1)if and only if (X0, X1)θ,1 → Y .

Hence we have Y ∈ Jθ(X0, X1) ∩Kθ(X0, X1) if and only if

(X0, X1)θ,1 → Y → (X0, X1)θ,∞.

In particular, (X0, X1)θ,p ∈ Jθ(X0, X1) ∩Kθ(X0, X1) for every p ∈ [1,∞].

Example 3.29 Ck(Rn) ∈ J1/2(Ck−1(Rn), Ck+1(Rn))∩K1/2(Ck−1(Rn), Ck+1(Rn))for all k ∈ N. More generally, Ck(Rn) ∈ J(k−m1)/(m2−m1)(C

m1(Rn), Cm2(Rn))∩K(k−m1)/(m2−m1)(C

m1(Rn), Cm2(Rn)) for all m1,m2, k ∈ N0 with m1 < k <m2.

Remark: It can be shown that C1(Rn) is not an interpolation space withrespect to (C(Rn), C2(Rn)), cf. [2, Example 1.3.3]!

The main result of this section is the following fundamental theorem:

Theorem 3.30 (Reiteration Theorem)Let (X0, X1) be an admissible pair of Banach spaces and let 0 ≤ θ0, θ1 ≤ 1 withθ0 6= θ1. Moreover, let θ ∈ (0, 1) and set ω = (1− θ)θ0 + θθ1. Then:

20

1. If Yj ∈ Kθj(X0, X1) for j = 0, 1, then

(Y0, Y1)θ,p → (X0, X1)ω,p for all 1 ≤ p ≤ ∞.

2. If Yj ∈ Jθj(X0, X1) for j = 0, 1, then

(X0, X1)ω,p → (Y0, Y1)θ,p for all 1 ≤ p ≤ ∞.

In particular, if Yj ∈ Jθj(X0, X1) ∩Kθj(X0, X1) for j = 0, 1, then

(Y0, Y1)θ,p = (X0, X1)ω,p for all 1 ≤ p ≤ ∞

with equivalent norms.

Remark 3.31 Since (X0, X1)θ,q ∈ Jθ(X0, X1)∩Kθ(X0, X1) for all θ ∈ (0, 1),1 ≤ q ≤ ∞, Theorem 3.30 yields

((X0, X1)θ0,q0 , (X0, X1)θ1,q1)θ,p = (X0, X1)(1−θ)θ0+θθ1,p

for all 0 < θ0 6= θ1 < 1 and 1 ≤ p, q0, q1 ≤ ∞. Moreover,

((X0, X1)θ0,q0 , X1)θ,p = (X0, X1)(1−θ)θ0+θ,p

(X0, (X0, X1)θ1,q1)θ,p = (X0, X1)(1−θ)+θθ1,p

for all 0 < θ0 6= θ1 < 1 and 1 ≤ p, q0, q1 ≤ ∞ since Xj ∈ Jj(X0, X1) ∩Kj(X0, X1) for j = 0, 1.

As consequences of the previous identity we obtain

(Cθ0(Rn), Cθ1(Rn))θ,∞ = (C0(Rn), C1(Rn))(1−θ)θ0+θθ1,∞

= C(1−θ)θ0+θθ1(Rn),

(W θ0p (Rn),W θ1

p (Rn))θ,p = (Lp(Rn),W 1p (Rn))(1−θ)θ0+θθ1,p

= W (1−θ)θ0+θθ1p (Rn)

for all 0 ≤ θ0 6= θ1 ≤ 1, θ ∈ (0, 1), and 1 ≤ p < ∞ due to Example 3.10.Moreover, since

C1(Rn) ∈ J 12(C0(Rn), C2(Rn)) ∩K 1

2(C0(Rn), C2(Rn)),

cf. Example 3.29, we have

(C0(Rn), C2(Rn))ω,∞ = (C0(Rn), C1(Rn))2ω,∞ = C2ω(Rn)

for all ω ∈ (0, 12).

In order to show (X0, X1)[θ] ∈ Kθ(X0, X1) we will need the followingintegral representation for bounded and holomorphic function on the strip:

21

Lemma 3.32 (Poisson formula)Let X be a Banach space and let f : S → X be continuous, bounded, andholomorphic in S0. Then f(z) = f0(z) + f1(z), where

fj(z) =

∫RPj(z, t)f(j + it) dt for all z ∈ S0, j = 0, 1, (3.9)

and

Pj(x+ iy, t) =eπ(y−t) sin(πx)

sin2(πx) + (cos(πx)− (−1)j exp(π(y − t)))2.

Proof: (Sketch) First let X = C, recall the Poisson formula for the unitcircle:

f(z) =1

2π

∫|w|=1

f(w)1− |z|2

|w − z|2dw for all |z| < 1,

which can be derive from the Cauchy formula, cf. e.g. Remmert: Funktio-nentheorie I. The representation on S can be derived from the latter Poissonformula with the aid of the mapping h : S → |z| ≤ 1 as in (2.2). Finally,the case that X is a general Banach space can be easily reduced to the scalarcase by considering z 7→ 〈f(z), x′〉 for an arbitrary x′ ∈ X ′.

Proposition 3.33 For every θ ∈ (0, 1) we have (X0, X1)[θ] ∈ Kθ(X0, X1).

As a consequence we obtain

THEOREM 3.34 Let 0 < θ0, θ1, θ < 1 with θ0 6= θ1 and let 1 ≤ q ≤ ∞.Then

((X0, X1)[θ0], (X0, X1)[θ1])θ,q = (X0, X1)(1−θ)θ0+θθ1,q

with equivalent norms.

For completeness we note that

((X0, X1)[θ0], (X0, X1)[θ1])[θ] = (X0, X1)[(1−θ)θ0+θθ1]

if X0 ∩ X1 is dense in X0, X1, and (X0, X1)[θ0] ∩ (X0, X1)[θ1], cf. [1, Theo-rem 4.6.1]. Note that, if X1 → X0, the latter conditions are valid.

Moreover, for all 0 < θ0 6= θ1 < 1, 0 < θ < 1, 1 ≤ p0, p1 ≤ ∞, and1p

= 1−θp0

+ θp1

we have

((X0, X1)θ0,p0 , (X0, X1)θ1,p1)[θ] = (X0, X1)(1−θ)θ0+θθ1,p,

cf. [1, Theorem 4.7.2].

22

3.5 Summary of Further Useful Results

3.5.1 Lorentz Spaces and Real Interpolation

In the following let (U, µ) be a measure space. Recall that for given measur-able f : U → K

m(t; f) = µ (x ∈ U : |f(x)| > t) , t ≥ 0,

is the distribution function of f .

Denition 3.35 The decreasing rearrangement of f is the function f ∗ : (0,∞)→[0,∞] dened by

f ∗(t) = inf σ > 0 : m(σ; f) ≤ t . (3.10)

Then f ∗ : (0,∞) → [0,∞) is decreasing and continuous from the right. Afundamental property of f ∗ is:

Lemma 3.36

m(ρ; f ∗) = m(ρ, f) for all ρ ≥ 0. (3.11)

Proof: First of all, (3.10) implies f ∗(m(ρ; f)) ≤ ρ. Since f ∗ is decreasing,this implies m(ρ; f ∗) ≤ |[0,m(ρ; f)]| = m(ρ; f). In order to prove the con-verse inequality, we note that f ∗(m(ρ; f ∗)) ≤ ρ since f ∗ is continuous fromthe right. Hence m(ρ; f) ≤ m(ρ, f ∗) due to (3.10).

Because of the latter lemma and [3, Theorem 8.16], we obtain∫U

|f(x)|p dµ(x) = p

∫ ∞0

tp−1m(t; f) dt

= p

∫ ∞0

tp−1m(t; f ∗) dt =

∫ ∞0

|f ∗(t)|p dt

for all 1 ≤ p <∞ and f ∈ Lp(U, µ). Moreover,

‖f‖L∞(U,µ) = inft > 0 : m(t; f) = 0 = ‖f ∗‖L∞(0,∞).

Denition 3.37 Let 1 ≤ p, q ≤ ∞. Then the Lorentz space Lp,q(U, µ) isdened as

Lp,q(U, µ) = f : U → K measurable : ‖f‖Lp,q <∞

‖f‖Lp,q =∥∥∥t 1pf ∗(t)∥∥∥

Lq(0,∞; dtt

).

23

As a direct consequence of the denition we obtain

Lp,p(U, µ) = Lp(U, µ), Lp,∞(U, µ) = Lpweak(U, µ)

for all 1 ≤ p ≤ ∞. Moreover, since f ∗ is monotone, one obtains

Lp,1(U, µ) → Lp,q1(U, µ) → Lp,q2(U, µ) → Lp,∞(U, µ) (3.12)

for every 1 ≤ q1 ≤ q2 ≤ ∞ similarly as in the proof of Lemma 3.6.The Lorentz spaces occur naturally as real interpolation spaces of Lp-

spaces:

THEOREM 3.38 Let 1 ≤ p0, p1, q0, q1, r ≤ ∞. Then for every θ ∈ (0, 1)we have

(Lp0,q0(U, µ), Lp1,q1(U, µ))θ,r = Lp,r(U, µ)

with equivalent norms if p0 6= p1, where1p

= 1−θp0

+ θp1. If p0 = p1, the same

conclusion holds if additionally 1q

= 1−θq0

+ θq1.

In the case 1 < p0 6= p1 < ∞, the latter theorem follows the ReiterationTheorem 3.30 and the following special case:

THEOREM 3.39 For every θ ∈ (0, 1)

(L1(U, µ), L∞(U, µ))θ,r = L1

1−θ ,r(U, µ).

The proof is based on the identity

K(t, f ;L1, L∞) =

∫ t

0

f ∗(τ) dτ for all t > 0

the monotonicity of f ∗(t) and the Hardy inequality. We refer to [2] for thedetails.

For the general case, a proof of Theorem 3.38 can be found in [1, Theo-rem 5.3.1].

3.5.2 Dual Spaces

THEOREM 3.40 Let (X0, X1) be an admissible couple of Banach spacessuch that X0 ∩ X1 is dense in X0 and X1 and let 1 ≤ p < ∞, θ ∈ (0, 1).Then

((X0, X1)θ,p)′ = (X ′0, X

′1)θ,p′ ,

((X0, X1)θ)′ = (X ′0, X

′1)θ,1.

24

Proof: See [2, Theorem 1.18].

We note that, if X0 ∩X1 is dense in X0 and X1, then

(X0 ∩X1)′ = X ′0 +X ′1, (X0 +X1)′ = X ′0 ∩X ′1

with same norms, cf. [1, Theorem 2.7.1]. Therefore ((X0, X1)θ,p)′ and (X ′0, X

′1)θ,p′

are intermediate spaces with respect to (X ′0, X′1) and the equalities above

make sense.

THEOREM 3.41 Let (X0, X1) be an admissible couple of Banach spacessuch that X0 ∩ X1 is dense in X0 and X1 and let θ ∈ (0, 1). Moreover, letX0 or X1 be reexive. Then(

(X0, X1)[θ]

)′= (X ′0, X

′1)[θ]

Proof: See [1, Corollary 4.5.2].

25

4 Bessel Potential and Besov Spaces

4.1 Mikhlin Multiplier Theorem

Recall that the Fourier transformation F and the inverse Fourier transfor-mation F−1 are dened by

F [f ](ξ) := f(ξ) :=

∫Rne−ix·ξf(x) dx,

F−1[f ](x) :=1

(2π)n

∫Rneix·ξf(ξ) dξ,

where f ∈ L1(Rn). We note that the denitions are the same for f ∈L1(Rn;X), where X is an arbitrary Banach space. Moreover, recall that byPlancharel's theorem F : L2(Rn) → L2(Rn) is an isomorphism with inverseF−1 and that ∫

Rnf(x)g(x) dx =

1

(2π)n

∫Rnf(ξ)g(ξ) dξ (4.1)

for all f, g ∈ L2(Rn). Here (4.1) holds true for f ∈ L2(Rn;X), g ∈ L2(Rn;X ′)if the products of the functions are understood as duality product pointwise.The proof is the same as in the scalar case or one can proof (4.1) in thevector-valued case by approximating f ∈ L2(Rn;X), g ∈ L2(Rn;X ′) bysimple functions and applying the identity in the scalar case. Furthermore, iff, g ∈ L2(Rn;H) and H is a Hilbert space, then F : L2(Rn;H)→ L2(Rn;H)is an isomorphism with inverse F−1.

THEOREM 4.1 (Mikhlin Multiplier Theorem)Let H0, H1 be Hilbert spaces and let N = n+2. Moreover, let m : Rn \0 →L(H0, H1) be an N-times dierentiable function such that

‖∂αξm(ξ)‖L(H0,H1) ≤ C|ξ|−|α| (4.2)

for all ξ 6= 0 and |α| ≤ N and let

m(Dx)f = F−1[m(ξ)f(ξ)

]for all f ∈ C∞0 (Rn;H0).

Then m(Dx) extends to a bounded linear operator

m(Dx) : Lp(Rn;H0)→ Lp(Rn;H1) for all 1 < p <∞. (4.3)

26

The theorem is proved in Appendix B. We just note that (4.1) implies that,if H0 = H1 = C,

‖m(Dx)f‖L2(Rn) ≤ ‖m‖L∞(Rn)‖f‖L2(Rn) for all f ∈ L2(Rn).

Hence m(Dx) ∈ L(L2(Rn)). In order to prove Theorem 4.1, the main step isto show that m(Dx) ∈ L(L1(Rn), L1

weak(Rn)). Once this is proved, m(Dx) ∈L(Lp(Rn)) for all 1 < p ≤ 2 by the Marcinkiewicz interpolation theoremand the case 2 < p < ∞ will follow by duality. Using that the PlancharelTheorem holds also for f ∈ L2(Rn;Hj), the proof can be generalized to thecase of general Hilbert spaces H0, H1.

4.2 A Fourier-Analytic Characterization of Hölder con-

tinuity

First of all, we recall that:

1. If f : Rn → C is a continuously dierentiable function such that f ∈L1(Rn) and ∂jf ∈ L1(Rn), then

F [∂xjf ] = iξjF [f ] = iξj f(ξ). (4.4)

2. If f ∈ L1(Rn) such that xjf ∈ L1(Rn), then f(ξ) is continuously partialdierentiable with respect to ξj and

∂ξj f(ξ) = F [−ixjf(x)]. (4.5)

The latter identities show the duality between dierentiability of f and decayof f(ξ) as |ξ| → ∞ as well as decay of f for large x and dierentiability of f .

In the following we will use a Littlewood-Paley partition of unity ϕj(ξ),j ∈ N0 (on Rn). This is a partition of unity ϕj(ξ), j ∈ N0, on Rn withϕj ∈ C∞0 (Rn) such that

suppϕj ⊆ξ ∈ Rn : 2j−1 ≤ |ξ| ≤ 2j+1

for all j ≥ 1. (4.6)

The partition of unity can be constructed such that suppϕ0 ⊂ B2(0), ϕj(ξ) =ϕ1(2−j+1ξ) for all j ≥ 1 and (4.6) holds. Then we have

|∂αξ ϕj(ξ)| ≤ C‖∂αξ ϕ1‖L∞(Rn)2−|α|j for all α ∈ Nn

0 , j ≥ 1. (4.7)

27

Moreover, we note that

ϕj(Dx)f = F−1[ϕj(ξ)f(ξ)

]= ϕj ∗ f

for all f ∈ C∞0 (Rn), j ∈ N0, where ϕj = F−1[ϕj] and

ϕj(x) = 2(j−1)nϕ1(2j−1x) for all j ∈ N and x ∈ Rn.

For f ∈ L∞(Rn) we will dene ϕj(Dx)f by

ϕj(Dx)f = ϕj ∗ f.

Furthermore, since ϕj−1 + ϕj + ϕj+1 ≡ 1 on suppϕj (where ϕ−1 ≡ 0), wehave

ϕj(Dx)f = (ϕj−1(Dx) + ϕj(Dx) + ϕj+1(Dx))ϕj(Dx)f (4.8)

for all f ∈ S(Rn), j ∈ N0.Using this decomposition, we obtain the following characterization of

Hölder continuous functions.

THEOREM 4.2 Let 0 < s < 1. Then f ∈ Cs(Rn) if and only if f ∈L∞(Rn) and

‖f‖Cs∗ := supj∈N0

2js‖ϕj ∗ f‖L∞(Rn) <∞.

Moreover, ‖ · ‖Cs∗ is an equivalent norm on Cs(Rn).

Proof: First let f ∈ Cs(Rn). Then

supx∈Rn|f(x− y)− f(x)| ≤ ‖f‖Cs|y|s

for all y ∈ Rn. Since we have chosen ϕj such that ϕj(ξ) = ϕ1(21−jξ) forj ≥ 1, we have ϕj(x) = ψ2−j(x) = 2jnψ(2jx), where ψ = F−1[ϕ1(2·)]. Thisimplies that

‖ϕj‖L1(Rn) ≤ C, ‖∇ϕj‖L1(Rn) ≤ C2j for all j ∈ N0. (4.9)

Moreover, ∫Rnϕj(y)dy =

∫Rnψ(y)dy = F [ψ](0) = ϕ1(0) = 0

28

for all j ≥ 1. Hence

ϕj(Dx)f =

∫Rnf(x− y)ψ2−j(y)dy =

∫Rn

(f(x− y)− f(x))ψ2−j(y)dy (4.10)

and therefore

‖ϕj(Dx)f‖∞ ≤ ‖f‖Cs∫Rn|y|sψ2−j(y)dy

= 2−js‖f‖Cs∫Rn|z|sψ(z)dz = C2−js‖f‖Cs

for all j ∈ N. The latter inequality implies ‖f‖Cs∗ ≤ C‖f‖Cs since also‖ϕ0(Dx)f‖∞ ≤ C‖f‖∞.

Conversely, let f ∈ L∞(Rn) be such that ‖f‖Cs∗ <∞. Now, if |y| ≤ 1,

f(x− y)− f(x) =∑

2j≤|y|−1

(fj(x− y)− fj(x)) +∑

2j>|y|−1

(fj(x− y)− fj(x)) ,

where fj = ϕj(Dx)f . In order to estimate the rst sum, we use the meanvalue theorem to conclude that

|fj(x− y)− fj(x)| ≤ |y|‖∇fj‖∞. (4.11)

Moreover, since

∂xkfj = ∂xkϕj−1(Dx)fj + ∂xkϕj(Dx)fj + ∂xkϕj+1(Dx)fj,

due to (4.8) and

‖∂xkϕl(Dx)g‖L∞(Rn) ≤ ‖∂xkϕl‖L1(Rn)‖g‖L∞(Rn) ≤ C2l‖g‖L∞(Rn)

for general l ∈ N0, g ∈ L∞(Rn), we obtain∑2j≤|y|−1

|fj(x− y)− fj(y)| ≤ C∑

2j≤|y|−1

|y|‖∇fj‖∞

≤ C|y|∑

2j≤|y|−1

2j(1−s)‖f‖Cs∗ ≤ C|y|s‖f‖Cs∗ .

The second sum is simply estimated by∑2j>|y|−1

|fj(x− y)− fj(y)| ≤ 2∑

2j>|y|−1

‖fj‖∞

≤ 2‖f‖Cs∗∑

2j>|y|−1

2−js = C|y|s‖f‖Cs∗

29

Altogether ‖f‖Cs ≤ C‖f‖Cs∗ .

Remark 4.3 Because of (4.9) we get

‖ϕj(Dx)f‖Lp(Rn) ≤ ‖ϕj‖L1(Rn)‖f‖Lp(Rn) ≤ ‖f‖Lp(Rn), (4.12)

‖∇ϕj(Dx)f‖Lp(Rn) ≤ ‖∇ϕj‖L1(Rn)‖f‖Lp(Rn) ≤ C2j‖f‖Lp(Rn) (4.13)

for any f ∈ Lp(Rn), 1 ≤ p ≤ ∞, j ∈ N0.

4.3 Bessel Potential and Besov Spaces Denitions and

Basic Properties

In the following let 〈ξ〉 := (1+|ξ|2)12 . We note that for any s ∈ R the function

〈ξ〉s is a smooth function satisfying

|∂αξ 〈ξ〉s| ≤ Cs,α(1 + |ξ|)s−|α| (4.14)

for all α ∈ Nn0 and some Cs,α > 0. The latter estimate can be proved by

considering f : Rn+1 \ 0 → R dened by fs(ξ, t) = |(t, ξ)|s. Since fs issmooth and homogeneous of degree s in (t, ξ), we have

|∂αξ fs(ξ, t)| ≤ Cs,α(|t|+ |ξ|)s−|α|

uniformly in (ξ, t) 6= 0 and for all α ∈ Nn0 .

Using (4.14) it is easy to show, that 〈ξ〉sf(ξ) ∈ S(Rn) for all f ∈ S(Rn).By duality 〈ξ〉sf ∈ S ′(Rn) for all f ∈ S ′(Rn). Therefore we can dene〈Dx〉s : S ′(Rn)→ S ′(Rn) as

〈Dx〉sf = F−1[〈ξ〉sf

]for all f ∈ S ′(Rn).

Denition 4.4 Let s ∈ R and let 1 < p <∞. Then the Bessel potential spaceHsp(Rn) is dened by

Hsp(Rn) = f ∈ S ′(Rn) : 〈Dx〉sf ∈ Lp(Rn)

‖f‖Hsp(Rn) = ‖〈Dx〉sf‖Lp(Rn) .

Remarks 4.5 1. If p = 2, then we have f ∈ Hs2(Rn) if and only if 〈ξ〉sf ∈

L2(Rn) by Plancharel's theorem.

30

2. By denition 〈Dx〉s : Hsp(Rn)→ Lp(Rn) is an isomorphism with inverse

〈Dx〉−s. Since S(Rn) is dense in Lp(Rn) and 〈Dx〉−s : S(Rn)→ S(Rn),S(Rn) is dense in Hs

p(Rn) for any s ∈ R, 1 < p <∞.

As a consequence of the Mikhlin multiplier theorem we obtain

Theorem 4.6 Let m ∈ N0 and let 1 < p < ∞. Then Hmp (Rn) = Wm

p (Rn)with equivalent norms.

Proof: We rst prove the embedding Hmp (Rn) → Wm

p (Rn). Let f ∈ S(Rn).Then

∂βxf = F−1[(iξ)β f(ξ)

]= F−1

[(iξ)β

〈ξ〉m〈ξ〉mf(ξ)

]Hence in order to obtain

‖∂βxf‖Lp(Rn) ≤ Cp‖〈Dx〉mf‖Lp(Rn) ≡ Cp‖f‖Hmp (Rn) (4.15)

for β ∈ Nn0 with |β| ≤ m we apply Theorem 4.1 to mβ(ξ) = (iξ)β

〈ξ〉m . Therefore

we have to verify (4.2) for m = mβ. To this end, we use (4.14) and

|∂αξ (iξ)β| ≤ Cα,β|ξ||β|−|α|. (4.16)

Moreover, (1 + |ξ|)−m−|α| ≤ |ξ|−|β|−|α| if |β| ≤ m. Therefore

|∂αξmβ(ξ)| ≤ Cα,β|ξ|−|α| (4.17)

follows from (4.16), (4.14), and the following claim:

Claim: Let s1, s2 ∈ R, N ∈ N and let p1, p2 : Rn \ 0 → C be N -timescontinuously dierentiable satisfying

|∂αξ pj(ξ)| ≤ C|ξ|sj−|α| for all |α| ≤ N, j = 1, 2.

Then|∂αξ (p1(ξ)p2(ξ))| ≤ C ′|ξ|s1+s2−|α| (4.18)

for all |α| ≤ N .

Proof of Claim: The claim follows directly from the Leibniz's formula.

Because of (4.17), the conditions of the Mikhlin Multiplier Theorem 4.1 aresatised and (4.15) follows for all |β| ≤ m, which proves Hm

p (Rn) → Wmp (Rn)

since S(Rn) is dense in Hmp (Rn).

31

Hence it remains to prove Wmp (Rn) → Hm

p (Rn). If m = 2k, k ∈ N0, iseven, then 〈ξ〉m = (1 + |ξ|2)k is a polynomial of degree m. Therefore 〈Dx〉mis a dierential operator of order m and

‖〈Dx〉mf‖Lp(Rn;H) ≤ C∑|α|≤m

‖∂αx f‖Lp(Rn;H),

which proves the embedding in this case.If m = 2k + 1, k ∈ N0, is odd, then

〈ξ〉m = 〈ξ〉m(

1

〈ξ〉2+

n∑j=1

ξ2j

〈ξ〉2

)=

1

〈ξ〉〈ξ〉2k +

n∑j=1

ξj〈ξ〉〈ξ〉2kξj,

where 〈ξ〉2k and 〈ξ〉2kξj are polynomials of degree at most 2k + 1. Hence

‖〈Dx〉mf‖Lp(Rn;H) ≤ C∑|α|≤m

n∑j=0

‖mj(Dx)∂αx f‖Lp(Rn;H),

where m0(ξ) = 〈ξ〉−1 and mj(ξ) = ξj〈ξ〉−1, j = 1, . . . , n. Hence it remainsto verify the Mikhlin condition (4.2) for mj(ξ). If j = 0, then (4.2) form(ξ) = m0(ξ) follows from (4.14) with s = −1 because of 〈ξ〉−1−|α| ≤ |ξ|−|α|.If j = 1, . . . , n, then (4.2) follows for m(ξ) = mj(ξ) from (4.14) with s = −1,(4.16) with β = ej, and (4.18).

Theorem 4.2 gives a motivation for the following denition of the Besovspace Bs

pq(Rn).

Denition 4.7 Let s ∈ R, 1 ≤ p, q ≤ ∞. Then

Bspq(Rn) =

f ∈ S ′(Rn) : ‖f‖Bspq(Rn) <∞

,

where

‖f‖Bspq(Rn) =

(∞∑j=0

2jsq‖ϕj(Dx)f‖qLp(Rn)

) 1q

if q <∞,

‖f‖Bspq(Rn) = supj∈N0

2js‖ϕj(Dx)f‖Lp(Rn) if q =∞.

32

Remarks 4.8 1. Because of Theorem 4.2, Cs(Rn) = Bs∞∞(Rn) for 0 <

s < 1. More generally, Cs∗(Rn) := Bs

∞∞(Rn), s > 0, are called Hölder-Zygmund spaces.

2. Note that f ∈ Bspq(Rn) if and only if

(ϕj(Dx)f)j∈N0 ∈ `sq(N0;Lp(Rn)),

Here `sq(M;X), M ⊆ Z, is the space of all X-valued sequences x =(xj)j∈M such that

‖x‖`sq(M;X) =

(∑

j∈M(2js‖xj‖X)q) 1q

if q <∞,supj∈M 2js‖xj‖X if q =∞.

Moreover, we set `q(M;X) = `0q(M;X). Of course (xj)j∈M ∈ `sq(M;X)

if and only if (2jsxj)j∈M ∈ `q(M;X).

3. Using Plancharel's theorem, it is not dicult to show that

Bs22(Rn) = Hs

2(Rn).

The proof is left to the reader as an exercise. But the statement willalso follow from Corollary 4.15 below.

Some simple relations between the Besov spaces are summarized in the fol-lowing:

Lemma 4.9 Let s ∈ R, 1 ≤ p, q1, q2 ≤ ∞, and let ε > 0. Then

Bspq1

(Rn) → Bspq2

(Rn) if q1 ≤ q2, Bs+εp∞ (Rn) → Bs

p1(Rn).

Proof: The rst embedding follows from

`q1(N0;X) → `q2(N0;X), `sq1(N0;X) → `sq2(N0;X) if q1 ≤ q2. (4.19)

The second embedding follows `s+ε∞ (N0;X) → `s1(N0;X) because of

‖(aj)‖`sq(N0;X) =∞∑j=0

2sj‖aj‖X

≤

(∞∑j=0

2−εj

)supj∈N0

2(s+ε)j‖aj‖X ≤ Cε‖(aj)‖`s+ε∞ (N0;X)

33

Remark 4.10 The latter lemma shows that q measures regularity of f ona ner scale than s, meaning, if s > s′, then Bs

pq1(Rn) → Bs′

pq2(Rn) with

arbitrary 1 ≤ q1, q2 ≤ ∞.

Exercise 1 Show thatB

np

p1(Rn) → C0(Rn).

Hint: Use that ϕj(Dx)f = ϕj ∗ f , where ‖ϕj‖Lq(Rn) ≤ C2j nq′ uniformly in

j ∈ Z.

In order to get a sharp comparision of Besov and Bessel potential spaces weprove:

THEOREM 4.11 Let s ∈ R, 1 < p <∞. Then there are constants c, C >0 such that

c‖f‖Hsp(Rn) ≤

∥∥∥∥∥∥(∞∑j=0

22js|ϕj(Dx)f(x)|2) 1

2

∥∥∥∥∥∥Lp(Rn)

≤ C‖f‖Hsp(Rn)

for all f ∈ Hsp(Rn).

Remark 4.12 Because of the latter equivalent norm on Hsp(Rn), one denes

more generally the Triebel-Lizorkin space F spq(Rn), s ∈ R, 1 < p, q <∞, as

F spq(Rn) =

f ∈ S ′(Rn) : ‖f‖F spq(Rn) <∞

,

‖f‖F spq(Rn) =

∥∥∥∥∥∥(∞∑j=0

2qjs|ϕj(Dx)f(x)|q) 1

q

∥∥∥∥∥∥Lp(Rn)

.

Hence the latter theorem shows that Hsp(Rn) = F s

p2(Rn). Finally, note that

‖f‖F spq(Rn) = ‖(ϕj(Dx)f)j∈N‖Lp(Rn;`sq(N0)) .

Proof of Theorem 4.11: First we will show that ‖f‖F sp2(Rn) ≤ C‖f‖Hsp(Rn).

To this end we dene a mapping

Q : S(Rn) ⊂ Lp(Rn)→ Lp(Rn; `2(N0))

34

by

(Qg)(x) = (2jsϕj(Dx)〈Dx〉−sg(x))j∈N0 ∈ `2(N0) for all x ∈ Rn.

Then(Qg)(x) = F−1

ξ 7→x [m(ξ)g(ξ)]

where m(ξ) ∈ L(C, `2(N0)) is dened by

m(ξ)a = (2jsϕj(ξ)〈ξ〉−s)j∈N0a for all a ∈ C, ξ ∈ Rn.

In order to show that Q extends to a bounded operator

Q : Lp(Rn)→ Lp(Rn; `2(N0)) for all 1 < p <∞, (4.20)

we verify the condition for the Mikhlin multiplier theorem 4.1:

‖∂αξm(ξ)‖2L(C,`2(N0)) =

∞∑j=0

22js|∂αξ (ϕj(ξ)〈ξ〉−s)|2

≤ Cα,s22js〈ξ〉−2s−2|α|χsuppϕj(ξ) ≤ Cα,s|ξ|−2|α|

for all α ∈ Nn0 , where we have used that 2j−1 ≤ |ξ| ≤ 2j+1 on suppϕj if j ≥ 1

and|∂αξ (ϕj(ξ)〈ξ〉−s)| ≤ Cα,s〈ξ〉−s−|α|

uniformly in j ∈ N0, which follows from (4.7), (4.14), and the product rule.Hence (4.20) follows and therefore

‖f‖F sp2(Rn) = ‖Q〈Dx〉sf‖Lp(Rn;`q(N0)) ≤ C‖〈Dx〉sf‖Lp(Rn) ≡ C‖f‖Hsp(Rn).

Note that we have shown that

Q : Hsp(Rn)→ Lp(Rn; `s2(N0))

is bounded, where

Qf = (2−jsQ〈Dx〉sf)j∈N0 = (ϕj(Dx)f)j∈N0 . (4.21)

Conversely, we dene a mapping

R : S(Rn; `2(N0)) ⊂ Lp(Rn; `2(N0))→ Lp(Rn)

35

by

(Ra)(x) =∞∑j=0

2−jsϕj(Dx)〈Dx〉saj(x) for all x ∈ Rn, a ∈ S(Rn; `2(N0)).

Here ϕj(ξ) = ϕj−1(ξ) +ϕj(ξ) +ϕj+1(ξ), j ∈ N0, and ϕ−1(ξ) = 0. Note thatϕj(ξ)ϕj(ξ) = ϕj(ξ) since ϕj(ξ) = 1 on suppϕj. Then

(Ra)(x) = F−1ξ 7→x [m(ξ)aj(ξ)]

where m(ξ) ∈ L(`2(N0),C) is dened by

m(ξ)a =∞∑j=0

2−jsϕj(ξ)〈ξ〉saj for all (aj)j∈N0 ∈ `2(N0).

Similarly, as before

‖∂αξm(ξ)‖2L(`2(N0),C) ≤

∞∑j=0

2−2js|∂αξ (ϕj(ξ)〈ξ〉s)|2

≤ Cq

∞∑j=0

2−2js〈ξ〉2s−2|α|χ2j−2≤|ξ|≤2j+2 ≤ Cq|ξ|−2|α|,

where we have used that for each ξ ∈ Rn at most 5 terms in the sum above arenon-zero and that 2−2js ≤ C〈ξ〉−s on supp ϕj ⊆ 2j−2 ≤ |ξ| ≤ 2j+2. Hence,applying Theorem 4.1 once more, we obtain that R extends to a boundedoperator

R : Lp(Rn; `2(N0))→ Lp(Rn) for all 1 < p <∞.

Now we apply R to aj = 2jsϕj(Dx)f , j ∈ N0. Then

Ra =∞∑j=0

2−jsϕj(Dx)〈Dx〉s2jsϕj(Dx)f = 〈Dx〉sf

since∑∞

j=0 ϕj(Dx)ϕj(Dx) =∑∞

j=0 ϕj(Dx) = I. Thus

‖f‖Hsp(Rn) = ‖〈Dx〉sf‖Lp(Rn) = ‖Ra‖Lp(Rn)

≤ C∥∥(2jsϕj(Dx)f)j∈N0

∥∥Lp(Rn;`2(N0))

= C‖f‖F sp2(Rn),

36

which proves the lemma. Finally, we note that the previous estimates implythat

R : Lp(Rn; `s2(N0))→ Hsp(Rn)

is bounded, where

R(aj)j∈N0 := 〈Dx〉sR(2jsaj)j∈N0 =∞∑j=0

ϕj(Dx)aj (4.22)

and therefore RQ = I on Hsp(Rn).

Remark 4.13 Note that the latter proof shows that Hsp(Rn) is a retract of

Lp(Rn; `s2(N0)). In general:

Denition 4.14 A Banach space X is called a retract of a Banach space Yif there are bounded, linear operators R : Y → X and Q : X → Y such thatRQ = idX .

If R, Q are dened by (4.22) and (4.21), then RQ = I on Hsp(Rn). Note

that the mappings are independent of p and s. Moreover, using the samemappings R and Q as in the previous proof it is easy to show that Bs

pq(Rn)is a retract of `q(N0;Lp(Rn)).

Corollary 4.15 Let 1 < p <∞, s ∈ R. Then

Bspp(Rn) →Hs

p(Rn) → Bsp2(Rn) if 1 < p ≤ 2, (4.23)

Bsp2(Rn) →Hs

p(Rn) → Bspp(Rn) if 2 ≤ p <∞. (4.24)

In particular, Hs2(Rn) = Bs

22(Rn) for all s ∈ R.

Proof: The statements follows from Theorem 4.11, the embeddings

`q(N0;Lp(Rn)) → Lp(Rn; `q(N0)) if 1 ≤ q ≤ p ≤ ∞ (4.25)

Lp(N0; `q(Rn)) → `q(Rn;Lp(N0)) if 1 ≤ p ≤ q ≤ ∞ (4.26)

37

as well as from (4.19). Here (4.25) follows from

‖(fj)j∈N0‖Lp(Rn;`q(N0))

=

∫Rn

(∞∑j=0

|fj(x)|q) p

q

dx

1p

=

∥∥∥∥∥∞∑j=0

|fj(.)|q∥∥∥∥∥

1q

Lpq (Rn)

≤

(∞∑j=0

‖|fj(.)|q‖Lpq (Rn)

) 1q

=

(∞∑j=0

‖fj‖qLp(Rn)

) 1q

,

where we have used Minkowski's inequality. The inequality (4.26) is provedanalogously.

4.4 Interpolation of Vector-Valued Lp-Spaces, Bessel Po-

tential, and Besov Spaces

Recall thatHsp(Rn) andBs

pq(Rn) are retract of Lp(Rn; `s2(N0)) and `sq(N0;Lp(Rn))with same retraction and co-retractions, which are independent of p, s. Moreprecisely,

Q : Hsp(Rn)→ Lp(Rn; `s2(N0)),

Q : Bspq(Rn)→ `sq(N0;Lp(Rn))

and

R : Lp(Rn; `s2(N0))→ Hsp(Rn),

R : `sq(N0;Lp(Rn))→ Bspq(Rn)

are bounded linear operators satisfying RQ = I where

Qf = (ϕj(Dx)f)j∈N0 , f ∈ Hsp(Rn) ∪Bs

pq(Rn).

and

R(aj)j∈N0 :=∞∑j=0

ϕj(Dx)aj, (aj)j∈N0 ∈ Lp(Rn; `s2(N0)) ∪ `sq(N0;Lp(Rn)).

Generally we have:

38

Proposition 4.16 Let (X0, X1) and (Y0, Y1) be admissible Banach spaces andlet Q : X0 + X1 → Y0 + Y1, R : Y0 + Y1 → X0 + X1 be linear mappings suchthat Q ∈ L(Xj, Yj), R ∈ L(Yj, Xj) and RQx = x for all x ∈ Xj and j = 0, 1.Then

R(Y0, Y1)θ,p = (X0, X1)θ,p R(Y0, Y1)[θ] = (X0, X1)[θ]

with equivalent norms for all θ ∈ (0, 1), 1 ≤ p ≤ ∞, where R(Y0, Y1)θ,p isequipped with the quotient norm

‖x‖R(Y0,Y1)θ,p = infy∈(Y0,Y1)θ,p:Ry=x

‖y‖θ,p.

Proof: The proof of the proposition is similar to the proof of Example 3.11.

Because of the latter proposition, it is sucient to obtain interpolationresults for Lp(Rn; `s2(N0)) and `sq(N0;Lp(Rn)) to characterize the real andcomplex interpolation spaces of Hs

p(Rn) and Bspq(Rn).

We start with a result for the real interpolation method:

THEOREM 4.17 Let X be a Banach spaces, let s0 6= s1 ∈ R, 0 < θ < 1,and let 1 ≤ q0, q1 ≤ ∞. Then for every 1 ≤ q ≤ ∞

(`s0q0(Z;X), `s1q1(Z;X))θ,q = `sq(Z;X),

(`s0q0(N0;X), `s1q1(N0;X))θ,q = `sq(N0;X)

with equivalent norms, where s = (1− θ)s0 + θs1.

As a consequence we obtain

THEOREM 4.18 Let s0 6= s1 ∈ R, 0 < θ < 1 and let s = (1− θ)s0 + θs1.Then for all 1 ≤ p, q0, q1, q ≤ ∞

(Bs0pq0

(Rn), Bs1pq1

(Rn))θ,q = Bspq(Rn)

with equivalent norms. Moreover, for any 1 ≤ q ≤ ∞, 1 < p <∞ we have

(Hs0p (Rn), Hs1

p (Rn))θ,q = Bspq(Rn)

with equivalent norms.

39

Remark 4.19 Because of Example 3.10, we know that

(Lp(Rn),W 1p (Rn))θ,p = W θ

p (Rn)

for all 1 ≤ p <∞. On the other hand, we have by the previous theorem:

(Lp(Rn),W 1p (Rn))θ,p = (H0

p (Rn), H1p (Rn))θ,p = Bθ

pp(Rn)

if 1 < p < ∞, θ ∈ (0, 1). Hence Bspp(Rn) = W s

p (Rn) for all s ∈ (0, 1),1 < p <∞ and

‖f‖W sp (Rn) = ‖f‖Lp(Rn) +

(∫Rn×Rn

|f(x)− f(y)|p

|x− y|θp+nd(x, y)

) 1p

is an equivalent norm on Bspp(Rn). A more general statement of this kind

will be given below.

For the complex interpolation method we have:

THEOREM 4.20 Let 1 ≤ p0, p1, q0, q1 <∞, 0 < θ < 1, and let (U, µ), (V, ν)be two measure spaces. Moreover, let 1

p= 1−θ

p0+ θ

p1, 1q

= 1−θq0

+ θq1. Then

(Lp0(U ;Lq0(V )), Lp1(U ;Lq1(V )))[θ] = Lp(U ;Lq(V ))

with equal norms.

Remark 4.21 If (X0, X1) is an admissible pair of Banach spaces and 1 ≤p0, p1 <∞, 0 < θ < 1, then

(Lp0(U ;X0), Lp1(U ;X1))[θ] = Lp(U ; (X0, X1)[θ])

with equal norms, where 1p

= 1−θp0

+ θp1, cf. [1, Theorem 5.1.2].

Corollary 4.22 Let s0, s1 ∈ R, 0 < θ < 1, and let 1 ≤ q0, q1, p0, p1 < ∞.Moreover, let 1

p= 1−θ

p0+ θ

p1, 1q

= 1−θq0

+ θq1. Then

(`s0q0(N0;Lp0(Rn)), `s1q1(N0;Lp1(Rn)))[θ] = `sq(N0;Lp(Rn))

(Lp0(Rn; `s0q0(N0)), Lp1(Rn; `s1q1(N0)))[θ] = Lp(Rn; `sq(N0))

with equal norms, where s = (1− θ)s0 + θs1.

As an application we obtain:

40

THEOREM 4.23 Let s0, s1 ∈ R, 0 < θ < 1 and let s = (1 − θ)s0 + θs1.Then

(Hs0p0

(Rn), Hs1p1

(Rn))[θ] = Hsp(Rn)

with equal norms for any 1 < p0, p1 < ∞ and 1p

= 1−θp0

+ θp1. Moreover, for

any 1 ≤ q0, q1, p0, p1 <∞ we have

(Bs0p0q0

(Rn), Bs1p1q1

(Rn))[θ] = Bspq(Rn)

with equal norms where 1p

= 1−θp0

+ θp1

and 1q

= 1−θq0

+ θq1.

4.5 Sobolev Embeddings and Traces

We start with a Sobolev-type embedding theorem for Besov and Bessel po-tential spaces.

THEOREM 4.24 Let s, s1 ∈ R with s ≤ s1 and 1 ≤ p1 ≤ p ≤ ∞ such that

s− n

p≤ s1 −

n

p1

.

Then

Bs1p1q1

(Rn) → Bspq(Rn) for all 1 ≤ q1 ≤ q ≤ ∞, (4.27)

Hs1p1

(Rn) → Hsp(Rn). if 1 < p1 ≤ p <∞ (4.28)

Proof: See [1, Theorem 6.51].Remarks on the proof: The embedding (4.27) follows from:

ϕk(Dx)f = ϕk(Dx)ϕk(Dx)f

with ϕk(Dx) = ϕk−1(Dx) + ϕk(Dx) + ϕk+1(Dx), ϕ−1(Dx) := 0 as well as

ϕk(Dx)g = ψ2−k ∗ g,

which implies

‖ϕk(Dx)g‖Lp(Rn) ≤ ‖ψ2−k‖Lq(Rn)‖g‖Lp1 (Rn) ≤ C2−k n

q′ ‖g‖Lp1 (Rn),

where 1q′

= 1 − 1q

= 1p1− 1

p. Then (4.28) follows from (4.27) by a clever

interpolation.

41

THEOREM 4.25 Let 1 ≤ p <∞, 1 ≤ q ≤ ∞, s > 1pand let Tr f = f |xn=0

for any continuous f ∈ Hsp(Rn) ∪ Bs

pq(Rn). Then Tr can be extended to abounded linear operator

Tr: Bspq(Rn)→ B

s− 1p

pq (Rn),

Tr: Hsp(Rn)→ B

s− 1p

pp (Rn).

Proof: See [1].Remarks on the proof: If m = 1, then

Tr: Hmp (Rn) = Wm

p (Rn)→ Wm− 1

pp (Rn−1) = B

s− 1p

pp (Rn−1)

follows from the trace method as before. Using a ∈ Bm− 1

ppp (Rn−1) if and only

if ∂αx′a ∈ B1− 1

ppp (Rn−1) for all |α| ≤ m−1, the same follows for general m ∈ N.

Then the statement follows for s ≥ 1 by interpolation.

In the case 1p< s ≤ 1 one uses that B

1p

p1(R) → C0(R). Therefore

|Tr f(x′)| ≤ C‖f(x′, .)‖B

1pp1(R)

⇒ ‖Tr f‖Lp(Rn−1) ≤ C‖f‖Lp(Rn−1;B

1pp1(R))

≤ C ′‖f‖B

1pp1(Rn)

.

An iterated interpolation then yields the statement of the theorem for 1p<

s < 1.

4.6 Equivalent Norms

The following theorem is the direct generalization of Theorem 4.2 for generalBesov spaces Bs

pq(Rn) with 0 < s < 1.

THEOREM 4.26 Let 0 < s < 1 and let 1 ≤ p, q ≤ ∞. Then there areconstants c, C (depending on s, p, q) such that

c‖f‖Bspq(Rn) ≤ ‖f‖Lp(Rn) +

(∫ ∞0

ωp(t; f)q

tsqdt

t

) 1q

≤ C‖f‖Bspq(Rn) (4.29)

if q <∞ and

c‖f‖Bspq(Rn) ≤ ‖f‖Lp(Rn) + supt>0

ωp(t; f)q

tsqdt

t≤ C‖f‖Bspq(Rn) (4.30)

42

if q =∞, whereωp(t; f) = sup

|h|≤t‖f(.+ h)− f‖Lp(Rn)

is the Lp-modulus of continuity of f .

Remark 4.27 We refer to [1, Theorem 6.2.5] for a more general statementin the case s > 0.

Proof of Theorem 4.26: We will only prove the case q < ∞ since theproof in the case q =∞ is a simple variant of the proof of Theorem 4.2.

First of all, since t 7→ ωp(t; f) is a monotone increasing function and t isproportional to 2−j on [2−j+1, 2−j],∫ 1

0

ωp(t; f)q

tsqdt

t≤ C

∞∑j=0

2sjqωp(2−j; f)q

and∞∑j=0

2sjqωp(2−j; f)q ≤ C

∫ 2

0

ωp(t; f)q

tsqdt

t.

Moreover,∫ ∞1

ωp(t; f)q

tsqdt

t≤ 2q‖f‖qLp(Rn)

∫ ∞1

t−sq−1 dt = C‖f‖qLp(Rn)

since ωp(t; f) ≤ 2‖f‖Lp(Rn). Hence we can replace the middle term in (4.29)by

‖f‖Lp(Rn) +

(∞∑j=0

2sjqωp(2−j; f)

) 1q

.

First we prove the second inequality in (4.29). For f ∈ Bspq(Rn) we denote

fk = ϕk(Dx)f . Then

‖fk(.+ h)− fk‖Lp(Rn) ≤ |h|‖∇fk‖Lp(Rn)

due to (4.11) and therefore

ωp(t; fk) ≤ t‖∇fk‖Lp(Rn) = t‖∇ϕk(Dx)ϕk(Dx)fk‖Lp(Rn)

≤ Ct2k‖ϕk(Dx)fk‖Lp(Rn)

43

because of (4.13), where ϕk(Dx) = ϕk−1(Dx) + ϕk(Dx) + ϕk+1(Dx), k ∈ N0,and ϕ−1(Dx) = 0. On the other hand, ωp(t, fk) ≤ 2‖fk‖Lp(Rn) and f =∑∞

k=0 fk. Therefore

2sjωp(2−j; f) ≤ C

(∞∑j=0

2sj min(1, 2−j+k)‖ϕk(Dx)fk‖Lp(Rn)

)

≤ C

(∞∑j=0

2s(j−k) min(1, 2−j+k)2sk‖ϕk(Dx)fk‖Lp(Rn)

).

Now, dening aj = C2sj min(1, 2−j), j ∈ Z, bj = 2sj‖ϕk(Dx)fj‖Lp(Rn) if j ≥ 0and bj = 0 else, we see that 2sjωp(2

−j; f) ≤ (a ∗ b)j, where

(a ∗ b)j =∑k∈Z

aj−kbk

is the convolution of two sequences. Hence(∞∑j=0

2sjqωp(2−j; f)q

) 1q

≤ ‖a ∗ b‖`q(Z) ≤ ‖a‖`1(Z)‖b‖`q(Z) ≤ C‖f‖Bspq(Rn),

where a ∈ `1(Z) since s ∈ (0, 1). Here we have used the discrete version ofYoung's inequality ‖a ∗ b‖`r ≤ ‖a‖`1‖b‖`r , which can be proved in the sameway as for the usual convolution using Hölder's inequality.

In order to prove the rst inequality in (4.29), we use that

ϕj(Dx)f =

∫Rn

(f(x− 2−jz)− f(x))ψ(z) dz,

cf. (4.10). Therfore

‖ϕj(Dx)f‖Lp(Rn) ≤∫Rn‖f(.− 2−jz)− f‖Lp(Rn)|ψ(z)| dz

≤∫Rnωp(2

−j|z|; f)|ψ(z)| dz.

44

and (∞∑j=1

2sjq‖ϕj(Dx)f‖qLp(Rn)

) 1q

≤∫Rn

(∞∑j=1

2sjqωp(2−j|z|; f)q

) 1q

|ψ(z)| dz

≤ C

∫Rn

(∫ ∞0

ωp(t|z|; f)q

(|z|t)sqdt

t

) 1q

|z|s|ψ(z)| dz

= C

(∫ ∞0

ωp(t; f)q

tsqdt

t

) 1q∫Rn|z|s|ψ(z)| dz,

where we can estimate∑∞

j=0 2sjqωp(2−j|z|; f)q by the corresponding inte-

gral by the same arguments as in the beginning of the proof. Finally,‖ϕ0(Dx)f‖Lp(Rn) ≤ C‖f‖Lp(Rn), which nishes the proof.

45

A Proof of the Marcinkiewicz Interpolation The-

orem

Proof of Theorem 1.8 if p0 = q0 = 1, p1 = q1 = r, (U, µ) = (Rn, λn):We note that the assumptions imply that

λ(t;Tf) := |x : |Tf(x)| > t| ≤ Cq‖f‖q

tq(A.1)

for q = 1 and q = r if r <∞ and ‖Tf‖∞ ≤ C∞‖f‖∞ if r =∞.First we consider the case r < ∞. Let f ∈ Lp(Rn) and consider the

distribution function λ(t;Tf), t > 0, dened as above. For given t > 0 wedene f = f1 + f2 by

f1(x) =

f(x) if |f(x)| > t,

0 else.

Then f1 ∈ L1(Rn) and f2 ∈ Lr(Rn) since∫Rn|f1(x)| dx =

∫Rn|f1(x)|p|f1(x)|1−p dx ≤ Cp(t)‖f‖pp

and similarly ∫Rn|f2(x)|r dx ≤ Cr,p(t)‖f‖pp.

Now, since |Tf(x)| ≤ |Tf1(x)|+ |Tf2(x)|, we have

x : |Tf(x)| > t ⊆ x : |Tf1(x)| > t/2 ∪ x : |Tf2(x)| > t/2 .

Therefore

λ(t;Tf) ≤ λ(t/2;Tf1) + λ(t/2;Tf2)

≤ C1

t/2

∫Rn|f1(x)| dx+

Crr

(t/2)r

∫Rn|f2(x)|r dx

=2C1

t

∫|f(x)|>t

|f(x)| dx+2rCr

r

tr

∫|f(x)|≤t

|f(x)|r dx

where we have used the weak type (1, 1) and (r, r) estimate and (A.1). Nextwe use that ∫

Rn|Tf(x)|p dx = p

∫ ∞0

tp−1λ(t;Tf) dt,

46

cf. e.g. [3, Theorem 8.16]. We combine this with the estimate before. Tothis end we calculate∫ ∞

0

tp−1t−1

(∫|f |>t|f | dx

)dt =

∫Rn|f |∫ |f |

0

tp−2 dt dx

=1

p− 1

∫Rn|f ||f |p−1 dx

since p > 1 and similarly∫ ∞0

tp−1t−r(∫|f |≤t|f |r dx

)dt =

∫Rn|f |r

∫ ∞|f |

tp−1−r dt dx

=1

r − p

∫Rn|f |r|f |p−r dx

since p < r. Altogether

‖Tf‖p ≤ Cp‖f‖p for all f ∈ Lp(Rn).

Finally, if r =∞, we assume for simplicity that C∞ = 1. Otherwise replaceT by C−1

∞ T . Then we use the same splitting of f as before, but cut at heightt/2 instead of t. Hence |Tf2(x)| ≤ t

2since ‖T‖L(L∞(Rn)) ≤ 1. Therefore

x : |Tf(x)| > t ⊆ x : |Tf1(x)| > t/2

and λ(t, Tf) ≤ λ(t/2, T f1). The rest of the proof in this case is done asbefore having only the rst term.

47

B Proof of the Mikhlin Multiplier Theorem

For simplicity we will rst prove the theorem in the case H0 = H1 = C andtherefore m : Rn \0 → L(C) ∼= C. How to modify the proof for the generalcase will be discussed below.

Note that (4.1) implies that

‖m(Dx)f‖L2(Rn) ≤ ‖m‖L∞(Rn)‖f‖L2(Rn) for all f ∈ L2(Rn).

Hence m(Dx) ∈ L(L2(Rn)). In order to prove Theorem 4.1, the main stepis to show that m(Dx) ∈ L(L1(Rn), L1

weak(Rn)), which is the content of thenext lemma. Once this is proved, m(Dx) ∈ L(Lp(Rn)) for all 1 < p ≤ 2 bythe Marcinkiewicz interpolation theorem and the case 2 < p <∞ will followby duality.

Lemma B.1 Let M ≡ m(Dx) be as in Theorem 4.1. Then

|x ∈ Rn : |Mf(x)| > t| ≤C‖f‖L1(Rn)

tfor all t > 0, f ∈ L1(Rn) (B.1)

for some C > 0 independent of t > 0.

In order to prove Lemma B.1, an essential ingredient will be the followingkernel estimate:

Proposition B.2 Let m : Rn \ 0 → C be as in Theorem 4.1. Then there isa locally integrable, continuously dierentiable function k : Rn \ 0 → C suchthat

m(Dx)f(x) =

∫Rnk(x− y)f(y) dy for all x 6∈ supp f, (B.2)

for all f ∈ L2(Rn) with compact support which satises

|k(z)| ≤ C|z|−n |∇k(z)| ≤ C|z|−n−1 (B.3)

for all z 6= 0.

We postpone the proof of Proposition B.2 to the end of this section.

48

Corollary B.3 Let Q be a cube and a ∈ L1(Rn) with supp a ⊆ Q and∫Qa(x) dx = 0. Then there is a constant C > 0 independent of a and Q

such that ∫Rn\Q

|m(Dx)a(x)| dx ≤ C‖a‖1,

where Q = Q2√n denotes the cube with same center as Q and 2

√n times the

side-length of Q.

Proof: Let x0 denote the center of Q. Then x 6∈ Q = Q2√n and y ∈ Q

implies |x− x0| > 2|y − x0|. Therefore∫Rn\Q

|m(Dx)a(x)| dx =

∫Rn\Q

|k ∗ a(x)| dx

≤∫Q

∫|x−x0|>2|y−x0|

|k(x− y)− k(x− x0)| dx|a(y)| dy

≤ C

∫Q

|a(y)| dy,

since∫Qa(x) dx = 0 provided that∫

|x|>2|y||k(x− y)− k(x)|dx ≤ C for all y ∈ Rn. (B.4)

In order to prove the latter estimate, we use

k(x− y)− k(x) = −∫ 1

0

y · ∇k(x− ty) dt.

If |x| > 2|y|, then

|k(x− y)− k(x)| ≤ supt∈[0,1]

|∇xk(x− ty)||y| ≤ C|x|−n−1|y|

since |x− ty| ≥ 12|x| for all t ∈ [0, 1]. Hence∫

|x|>2|y||k(x− y)− k(x)| dx ≤ C

∫|x|>2|y|

|x|−n−1 dx|y| ≤ C ′

uniformly in y 6= 0.

49

Proposition B.4 Let f ∈ L1(Rn) be continuous and let t > 0. Then thereare cubes Qj, j ∈ N ⊆ N0, with disjoint interior and parallel to the axis suchthat

1.

t <1

|Qj|

∫Qj

|f(y)| dy ≤ 2nt for all j ∈ N. (B.5)

2. |f(x)| ≤ t for (almost) all x 6∈⋃j∈N Qj.

Proof: In the following Dk, k ∈ Z, denotes the set of all dyadic cubes withside length 2−k meaning the collection of all (closed) cubes Q with cornerson neighboring points of the lattice 2−kZn. Moreover, we set D =

⋃k∈ZDk.

Let C ′t for given t > 0 be the set of all Q ∈ D satisfying the condition

t <1

|Q|

∫Q

|f(x)| dx

and let Ct be the subset of all Q ∈ C ′t that are maximal with respect toinclusion in C ′t. Every Q ∈ C ′t is contained in some Q′ ∈ Ct since |Q| ≤ t−1‖f‖1

for all Q ∈ C ′t. Next, if Q ∈ Ct ∩ Dk and Q ⊂ Q′ ∈ Dk−1, then by themaximality of Q we have Q′ 6∈ C ′t, i.e.,

1

|Q′|

∫Q′|f(x)| dx ≤ t.

Moreover, since |Q′| = 2n|Q|, we get

t <1

|Q|

∫Q

|f(x)| dx ≤ 2n

|Q′|

∫Q′|f(x)| dx ≤ 2nt for all Q ∈ Ct.

Hence Ct = Qj : j ∈ N, where Qj, j ∈ N ⊆ N0, are non-overlapping and(B.5) is satised for all j ∈ N .

Now let F := Rn \⋃j∈N Qj. If x ∈ F , then 1

|Q|

∫Qf(y) dy ≤ t for every

Q ∈ D such that x ∈ Q. Hence, choosing a sequence of cubes Qk ∈ Dk withx ∈ Qk, we obtain

|f(x)| = limk→∞

∣∣∣∣ 1

|Qk|

∫Qk

f(y) dy

∣∣∣∣ ≤ t for all x ∈ F.

50

Remark B.5 The latter proposition holds for a general f ∈ L1(Rn) if oneapplies Lebesgue's dierentiation theorem in the last step of the proof.

Proof of Lemma B.1: First of all, since C∞0 (Rn) is dense in L1(Rn), itis enough to consider f ∈ C∞0 (Rn). Moreover, let t > 0 be xed and letQj, j ∈ N ⊆ N0 be the cubes due to Proposition B.4, Ω =

⋃j∈N Qj and

F = Rn \ Ω. We dene g, b ∈ L1(Rn) ∩ L2(Rn) by

g(x) =

f(x) if x ∈ F

1|Qj |

∫Qjf(y) dy if x ∈ Qj

and b(x) = f(x)− g(x). Note that this implies

1. |g(x)| ≤ 2nt almost every in Rn,

2. b(x) = 0 for every x ∈ F and∫Qjb(x) dx = 0 for each j ∈ N .

Then

|x : |Mf(x)| > t| ≤ |x : |Mg(x)| > t/2|+ |x : |Mb(x)| > t/2|

and it is sucient to estimate each term separately.In order to estimateMg, we use that |g(x)| ≤ 2nt for almost every x ∈ Rn,

f(x) = g(x) for x ∈ F , t|Ω| ≤ ‖f‖1, and that M ∈ L(L2(Rn)). Moreprecisely,

|x : |Mg(x)| > t/2| ≤ 4

t2

∫|Mg(x)|2 dx ≤ 4

t2‖m‖∞

∫|g(x)|2 dx

≤ C‖m‖∞t−2

(∫F

t|f(x)| dx+ t2|Ω|)

≤ Ct−1‖f‖1

In order to estimate Tb, we apply Corollary B.3 to bj(x) := b(x)χQj(x) andconclude ∫

Rn\Qj|Mbj(x)| dx ≤ C‖bj‖1 ≤ 2C

∫Qj

|f(x)| dx

where Qj = Q2√n

j . On the other hand, since b ∈ L2(Rn),∑

j∈N bj and

therefore∑

j∈N Mbj converge in L2(Rn) to b and Tb, respectively. Hence

|Mb(x)| ≤∑j∈N

|Mbj(x)| almost everywhere

51

and ∫Rn\Ω|Mb(x)| dx ≤ 2C

∑j∈N

∫Qj

|f(x)| dx ≤ 2C‖f‖1,

where Ω =⋃j∈N Qj. Finally,

|x : |Mb(x)| > t/2| ≤ |Ω|+ 2

t

∫Rn\Ω|Mb(x)| dx ≤ C

t‖f‖1,

where we have used that

|Ω| ≤∑j∈N

|Qj| ≤ (2√n)n

∑j∈N

|Qj| ≤C

t

∑j∈N

∫Qj

|f(x)| dx ≤ C

t‖f‖1

This nishes the proof of (B.1).

Proof of Theorem 4.1: Since ‖m(Dx)‖L(L2(Rn)) ≤ ‖m‖∞ and because of(B.1), we can apply the Marcinkiewicz interpolation theorem to concludem(Dx) ∈ L(Lp(Rn)) for all 1 < p ≤ 2. The statement for 2 < p <∞ followsby duality since∫

Rnm(Dx)f(x)g(x) dx =

1

(2π)n

∫Rnm(ξ)f(ξ)g(ξ) dξ

=1

(2π)n

∫Rnf(ξ)m(ξ)g(ξ) dξ

=

∫Rnf(x)m(Dx)g(x) dx

for all f, g ∈ C∞0 (Rn) due to (4.1), where m(ξ) = m(ξ) satises the sameestimates as m(ξ). Therefore m(Dx) = m(Dx)

′ ∈ L(Lp′(Rn)) if 2 < p < ∞

by the rst part.

It remains to prove Proposition B.2. To this end we will use the so-calledLittlewood-Paley partition of unity, which is nowadays a standard tool in thetheory of function spaces and harmonic analysis. It is frequently used toanalyze mapping properties of certain operators. Usually a Littlewood-Paleyor dyadic partition of unity on Rn \ 0 is a decomposition of unity ϕj(ξ),j ∈ Z of Rn \ 0 such that

suppϕj(ξ) ⊆ ξ ∈ Rn : c2j ≤ |ξ| ≤ C2j for all j ∈ Z.

52

Here c, C > 0 are some suitable xed numbers often chosen to be c = 12, C =

2, which we will assume in the following. Such a partition can be easilyconstructed by choosing some non-negative ψ ∈ C∞0 (Rn) such that ψ(ξ) > 0if and only if 1

2< |ξ| < 2. Then dening ψj(ξ) := ψ(2−jξ), j ∈ Z we have

Φ(ξ) =∑j∈Z

ψj(ξ) > 0 for all ξ 6= 0,

where we note that for each ξ 6= 0 the sum above contains at most twonon-vanishing terms. Hence

ϕj(ξ) = Φ(ξ)−1ψj(ξ)

denes a decomposition of unity with the desired properties. Moreover, inthis case ϕj(ξ) = ϕ0(2−jξ) since Φ(2−jξ) = Φ(ξ), which implies that

|∂αξ ϕj(ξ)| ≤ C‖∂αξ ϕ0‖L∞(Rn)2−|α|j (B.6)

for all α ∈ Nn0 and j ∈ Z.

The idea of the proof of Proposition B.2 is to decompose

m(ξ) =∑j∈Z

mj(ξ), ξ 6= 0,

where mj(ξ) = ϕj(ξ)m(ξ). Then

|∂αξmj(ξ)| ≤∑

0≤β≤α

(α

β

)|∂α−βξ ϕj(ξ)||∂βξm(ξ)| ≤ C2−j|α| (B.7)

because of (4.2), (B.6), and since 2−j−1 ≤ |ξ| ≤ 2−j+1 on suppϕj.For each part mj(ξ), we have

mj(Dx)f = F−1[mj(ξ)f(ξ)

]=

∫Rnkj(x− y)f(y) dy,

wherekj(x) = F−1[mj](x) ∈ C0(Rn)

since mj(ξ) ∈ L1(Rn). Here we have used that

F [f ∗ g](ξ) = f(ξ)g(ξ) for all ξ ∈ Rn, f, g ∈ L1(Rn).

53

Hence formally

m(Dx)f =∑j∈Z

∫Rnkj(x− y)f(y) dy,

where it remains to show that the sum on the right-hand side converges forx 6∈ supp f and that

k(z) =∑j∈Z

kj(z), z 6= 0,

converges to a function satisfying (B.3). To this end, we need some suitableuniform estimates of kj(z). These are a consequence of the following simplelemma:

Lemma B.6 Let N ∈ N0 and let g : Rn → C be an N-times dierentiablefunction, N ∈ N0, with compact support. Then

|F−1[g](x)| ≤ CN | supp g||x|−N sup|β|=N

‖∂βξ g‖L∞(Rn) (B.8)

uniformly in x 6= 0 and g, where CN depends only on n,N .

Proof: Let β ∈ Nn0 with |β| = N . Then

(−ix)βF−1[g] = F−1[∂βξ g]

and therefore

|xβ||F−1[g](x)| ≤ C‖∂βξ g‖L1(Rn) ≤ C| supp g| sup|β|=N

‖∂βξ g‖L∞(Rn).

Since β ∈ Nn0 with |β| = N was arbitrary,

|x|N |F−1[g](x)| ≤ CN | supp g| sup|β|=N

‖∂βξ g‖L∞(Rn)

for all x ∈ Rn, which completes the proof.

Corollary B.7 Let m be as in the assumptions of Theorem 4.1 and letmj(ξ) = m(ξ)ϕj(ξ), j ∈ Z, where ϕj, j ∈ Z, is the dyadic decompositionof unity of Rn \ 0 as above. Then kj(x) := F−1

ξ 7→x[mj] satises

|∂αz kj(z)| ≤ Cα,n,m2j(n+|α|−M)|z|−M (B.9)

for all z 6= 0, M = 0, . . . , n+2, α ∈ Nn0 for some constant Cα,n,m independent

of j ∈ Z, z 6= 0.

54

Proof: First of all,

∂αz kj(z) = F−1 [(iξ)αmj(ξ)] .

Using (B.7) and 2j−1 ≤ |ξ| ≤ 2j+1 on suppϕj, one easily veries

|∂βξ ((iξ)αmj(ξ)) | ≤ Cα,β2j(|α|−|β|)

for all |β| ≤ n + 2. Hence, applying Lemma B.6 to (iξ)αmj with N = M =0, . . . , n+ 2, we obtain

|∂αz kj(z)| ≤ Cn,m| suppϕj|2j(|α|−M)|z|−M ≤ C ′n,m2j(n+|α|−M)|z|−M ,

which nishes the proof.

Proof of Proposition B.2: Firstly, we will show that∑

j∈Z ∂αz kj(z), |α| ≤ 1

converges absolutely and uniformly on every compact subset of Rn \ 0 toa function ∂αz k(z) satisfying (B.3). The main idea of the proof is to split forgiven z 6= 0 the sum

∑j∈Z kj(z) into the two parts∑

2j≤|z|−1

∂αz kj(z) and∑

2j>|z|−1

∂αz kj(z)

and to show convergence and the estimate (B.3) separately.For the rst sum we use (B.9) with |α| ≤ 1 and M = 0. Then∑

j≤ld |z|−1

|∂αz kj(z)| ≤ C∑

j≤ld |z|−1

2j(n+|α|) ≤ C ′|z|−n−|α|

where ld denotes the logarithm with respect to basis 2. For the second sumwe apply (B.9) with |α| ≤ 1 and M = n+ |α|+ 1 and obtain∑

ld |z|−1<j

|∂αz kj(z)| ≤ 2∑

ld |z|−1<j

2−j|z|−n−|α|−1 ≤ C ′|z|−n−|α|.

Hence∑

j∈Z ∂αz kj(z) converges absolutely and uniformly on every closed sub-

set of Rn \ 0 to a function k(z) that satises (B.3) for all |α| ≤ 1.Finally, it remains to show that k(z) satises (B.2). First of all,

m(ξ)f(ξ) =∑j∈Z

mj(ξ)f(ξ), f ∈ C∞0 (Rn)

55

since∑

j∈Z ϕj(ξ) is locally nite. Moreover, since |mj(ξ)| is uniformly bounded

w.r.t. ξ and f(ξ) ∈ L2(Rn), the sum on the right-hand side converges inL2(Rn) to the left-hand side by Lebesgue's theorem on dominated conver-gence. Hence

m(Dx)f = limk→∞

∑|j|≤Nk

mj(Dx)f = limk→∞

∑|j|≤Nk

∫Rnkj(x− y)f(y) dy,

in L2(Rn) and almost everywhere for every f ∈ C∞0 (Rn) and some Nk →∞as k →∞ since Fourier transformation is bounded operator from L2(Rn) toL2(Rn). Therefore it only remains to interchange the summation and inte-gration in last term above provided that x 6∈ supp f . But this can be donesince

∑j∈Z kj(z) converges absolutely and uniformly on every closed subset

of Rn \ 0 as shown above. Hence (B.2) follows.

Comments on the proof of the vector-valued case: In the case thatm : Rn\0 → L(H0, H1) satises the assumptions of Theorem 4.1 for generalHilbert spaces H0, H1, we still conclude immediately that

m(Dx) : L2(Rn;H0)→ L2(Rn;H1)

is a bounded operator due Plancharel theorem.5 The rest of the proof canbe modied in a straight-forward manner to the operator-valued case sinceall arguments are based on direct estimates involving only the size of f(x),m(ξ), and k(z). One just has to replace the pointwise absolute value |.| bythe corresponding norms, i.e., ‖.‖H0 , ‖.‖H1 , and ‖.‖L(H0,H1). Finally, we notethat the Marcinkiewicz interpolation theorem holds for general X-valued Lp-spaces for general Banach spaces X since its proof is only based on argumentsinvolving the size of the function.

5Actually, this is the only step in the proof, where it is needed that H0, H1 are Hilbert

space and not general Banach spaces.

56

References

[1] J. Bergh and J. Löfström. Interpolation Spaces. Springer, Berlin - Hei-delberg - New York, 1976.

[2] Alessandra Lunardi. Interpolation Theory. Appunti. Scuola NormaleSuperiore, Pisa, 1999.

[3] W. Rudin. Real and complex analysis. McGraw-Hill Book Co., New York,third edition, 1987.

[4] Luc Tartar. An introduction to Sobolev spaces and interpolation spaces,volume 3 of Lecture Notes of the Unione Matematica Italiana. Springer,Berlin, 2007.

[5] H. Triebel. Interpolation Theory, Function Spaces, Dierential Opera-tors. North-Holland Publishing Company, Amsterdam, New York, Ox-ford, 1978.

57